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Finding the smallest integer that appears at least k times

You are given an array A of integers and an integer k. Implement an algorithm that determines, in linear time, the smallest integer that appears at least k times in A.
I have been struggling with this problem for awhile, coding in Java, I need to use a HashTable to find the smallest integer that appears at least k times, it also must be in linear time.
This is what I attempted but it does not pass any of the tests
private static int problem1(int[] arr, int k)
{
// Implement me!
HashMap<Integer, Integer> table = new HashMap<Integer, Integer>();
int ans = Integer.MAX_VALUE;
for (int i=0; i < arr.length; i++) {
if(table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
if (k <= table.get(arr[i])) {
ans = Math.min(ans, arr[i]);
}
}else{
table.put(arr[i], 1);
}
}
return ans;
}
Here is the empty code with all of the test cases:
import java.io.*;
import java.util.*;
public class Lab5
{
/**
* Problem 1: Find the smallest integer that appears at least k times.
*/
private static int problem1(int[] arr, int k)
{
// Implement me!
return 0;
}
/**
* Problem 2: Find two distinct indices i and j such that A[i] = A[j] and |i - j| <= k.
*/
private static int[] problem2(int[] arr, int k)
{
// Implement me!
int i = -1;
int j = -1;
return new int[] { i, j };
}
// ---------------------------------------------------------------------
// Do not change any of the code below!
private static final int LabNo = 5;
private static final String quarter = "Fall 2020";
private static final Random rng = new Random(123456);
private static boolean testProblem1(int[][] testCase)
{
int[] arr = testCase[0];
int k = testCase[1][0];
int answer = problem1(arr.clone(), k);
Arrays.sort(arr);
for (int i = 0, j = 0; i < arr.length; i = j)
{
for (; j < arr.length && arr[i] == arr[j]; j++) { }
if (j - i >= k)
{
return answer == arr[i];
}
}
return false; // Will never happen.
}
private static boolean testProblem2(int[][] testCase)
{
int[] arr = testCase[0];
int k = testCase[1][0];
int[] answer = problem2(arr.clone(), k);
if (answer == null || answer.length != 2)
{
return false;
}
Arrays.sort(answer);
// Check answer
int i = answer[0];
int j = answer[1];
return i != j
&& j - i <= k
&& i >= 0
&& j < arr.length
&& arr[i] == arr[j];
}
public static void main(String args[])
{
System.out.println("CS 302 -- " + quarter + " -- Lab " + LabNo);
testProblems(1);
testProblems(2);
}
private static void testProblems(int prob)
{
int noOfLines = prob == 1 ? 100000 : 500000;
System.out.println("-- -- -- -- --");
System.out.println(noOfLines + " test cases for problem " + prob + ".");
boolean passedAll = true;
for (int i = 1; i <= noOfLines; i++)
{
int[][] testCase = null;
boolean passed = false;
boolean exce = false;
try
{
switch (prob)
{
case 1:
testCase = createProblem1(i);
passed = testProblem1(testCase);
break;
case 2:
testCase = createProblem2(i);
passed = testProblem2(testCase);
break;
}
}
catch (Exception ex)
{
passed = false;
exce = true;
}
if (!passed)
{
System.out.println("Test " + i + " failed!" + (exce ? " (Exception)" : ""));
passedAll = false;
break;
}
}
if (passedAll)
{
System.out.println("All test passed.");
}
}
private static int[][] createProblem1(int testNo)
{
int size = rng.nextInt(Math.min(1000, testNo)) + 5;
int[] numbers = getRandomNumbers(size, size);
Arrays.sort(numbers);
int maxK = 0;
for (int i = 0, j = 0; i < size; i = j)
{
for (; j < size && numbers[i] == numbers[j]; j++) { }
maxK = Math.max(maxK, j - i);
}
int k = rng.nextInt(maxK) + 1;
shuffle(numbers);
return new int[][] { numbers, new int[] { k } };
}
private static int[][] createProblem2(int testNo)
{
int size = rng.nextInt(Math.min(1000, testNo)) + 5;
int[] numbers = getRandomNumbers(size, size);
int i = rng.nextInt(size);
int j = rng.nextInt(size - 1);
if (i <= j) j++;
numbers[i] = numbers[j];
return new int[][] { numbers, new int[] { Math.abs(i - j) } };
}
private static void shuffle(int[] arr)
{
for (int i = 0; i < arr.length - 1; i++)
{
int rndInd = rng.nextInt(arr.length - i) + i;
int tmp = arr[i];
arr[i] = arr[rndInd];
arr[rndInd] = tmp;
}
}
private static int[] getRandomNumbers(int range, int size)
{
int numbers[] = new int[size];
for (int i = 0; i < size; i++)
{
numbers[i] = rng.nextInt(2 * range) - range;
}
return numbers;
}
}

private static int problem1(int[] arr, int k) {
// Implement me!
Map<Integer, Integer> table = new TreeMap<Integer, Integer>();
for (int i = 0; i < arr.length; i++) {
if (table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
} else {
table.put(arr[i], 1);
}
}
for (Map.Entry<Integer,Integer> entry : table.entrySet()) {
//As treemap is sorted, we return the first key with value >=k.
if(entry.getValue()>=k)
return entry.getKey();
}
//Not found
return -1;
}

As others have pointed out, there are a few mistakes. First, the line where you initialize ans,
int ans = 0;
You should initialize ans to Integer.MAX_VALUE so that when you find an integer that appears at least k times for the first time that ans gets set to that integer appropriately. Second, in your for loop, there's no reason to skip the first element while iterating the array so i should be initialized to 0 instead of 1. Also, in that same line, you want to iterate through the entire array, and in your loop's condition right now you have i < k when k is not the length of the array. The length of the array is denoted by arr.length so the condition should instead be i < arr.length. Third, in this line,
if (k < table.get(arr[i])){
where you are trying to check if an integer has occurred at least k times in the array so far while iterating through the array, the < operator should be changed to <= since the keyword here is at least k times, not "more than k times". Fourth, k should never change so you can get rid of this line of code,
k = table.get(arr[i]);
After applying all of those changes, your function should look like this:
private static int problem1(int[] arr, int k)
{
// Implement me!
HashMap<Integer, Integer> table = new HashMap<Integer, Integer>();
int ans = Integer.MAX_VALUE;
for (int i=0; i < arr.length; i++) {
if(table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
if (k <= table.get(arr[i])) {
ans = Math.min(ans, arr[i]);
}
}else{
table.put(arr[i], 1);
}
}
return ans;
}

Pseudo code:
collect frequencies of each number in a Map<Integer, Integer> (number and its count)
set least to a large value
iterate over entries
ignore entry if its value is less than k
if entry key is less than current least, store it as least
return least
One line implementation:
private static int problem1(int[] arr, int k) {
return Arrays.stream(arr).boxed()
.collect(groupingBy(identity(), counting()))
.entrySet().stream()
.filter(entry -> entry.getValue() >= k)
.map(Map.Entry::getKey)
.reduce(MAX_VALUE, Math::min);
}

This was able to pass all the cases! Thank you to everyone who helped!!
private static int problem1(int[] arr, int k)
{
// Implement me!
HashMap<Integer, Integer> table = new HashMap<Integer, Integer>();
int ans = Integer.MAX_VALUE;
for (int i=0; i < arr.length; i++) {
if(table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
}else{
table.put(arr[i], 1);
}
}
Set<Integer> keys = table.keySet();
for(int i : keys){
if(table.get(i) >= k){
ans = Math.min(ans,i);
}
}
if(ans != Integer.MAX_VALUE){
return ans;
}else{
return 0;
}
}

I made a MergeSort method with the firstHalf and secondHalf Method; index 0 ~ middle, middle+1 ~ end. However, this calls the StackOverFlow Error

public static void mergeSort(int[] data) {
int[] left = firstHalf(data);
int[] right = secondHalf(data);
if (data.length > 1) {
mergeSort(left);
mergeSort(right);
merge(data, left, right);
}
}
public static void merge(int[] data, int[] left, int[] right) {
int tempArraySize = data.length;
int mergedNumbers[] = new int[tempArraySize]; //Temp array to take the sorted array
int mergePos;
int leftPos;
int rightPos;
int middle = data.length / 2;
mergePos = 0;
leftPos = 0; // 0 index
rightPos = middle + 1; //j is middle index
while (leftPos <= middle && rightPos <= data.length - 1) {
if (left[leftPos] < right[rightPos]) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
} else {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
}
mergePos++;
}
// when the right half array finishes sorting
while (leftPos <= middle) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
}
// when the left half array finishes sorting
while (rightPos <= data.length - 1) {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
// give value to the original array
for (mergePos = 0; mergePos < tempArraySize; ++mergePos) {
data[leftPos + mergePos] = mergedNumbers[mergePos];
}
}
public static int[] firstHalf(int[] data) {
int[] tempFirst = new int[(data.length / 2) + 1];
for (int i = 0; i <= data.length / 2; i++) {
tempFirst[i] = data[i];
}
return tempFirst;
}
public static int[] secondHalf(int[] data) {
int[] tempSecond = new int[(data.length / 2) + 1];
for (int i = (data.length / 2) + 1; i < data.length; i++) { // Middle to the end
for (int j = 0; j <= data.length / 2; j++) {
tempSecond[j] = data[i];
}
}
return tempSecond;
}
This is what I made.
My mergeSort method makes an error java.lang.StackOverflowError
What mistakes I made?
I made the firstHalf and secondHalf methods to get the index 0 ~ middle and middle+1 ~ end.
Those methods are made to get the value from the original 'data' Array.
The merge method is as same as the common MergeSort code.
Do I have to build a base case in the mergeSort method?

With this approach, it is simpler to return merged arrays. It would be faster to do a one time allocation of a temporary array, and use indexing to merge data between the two arrays rather than creating temporary arrays and copy data. Fixes noted in comments.
public static int[] mergeSort(int[] data) { // fix
int[] left = firstHalf(data);
int[] right = secondHalf(data);
if(data.length < 2) // change
return data; // fix
left = mergeSort(left); // fix
right = mergeSort(right); // fix
return merge(left, right); // fix
}
public static int[] merge(int[] left, int[] right) { // fix
int mergedNumbers [] = new int[left.length+right.length]; // fix
int mergePos = 0; // fix
int leftPos = 0; // fix
int rightPos = 0; // fix
while (leftPos < left.length && rightPos < right.length) { // fix
if (left[leftPos] < right[rightPos]) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
} else {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
}
mergePos++;
}
while (leftPos < left.length) { // fix
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
}
while (rightPos < right.length) { // fix
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
return mergedNumbers; // fix
}
public static int[] firstHalf(int[] data) {
int j = (data.length/2); // fix
int[] tempFirst = new int[j]; // fix
for(int i = 0; i < tempFirst.length; i++) // fix
tempFirst[i] = data[i];
return tempFirst;
}
public static int[] secondHalf(int[] data) {
int j = (data.length/2); // fix
int[] tempSecond = new int[data.length-j]; // fix
for(int i = 0; i < tempSecond.length; i++) // fix
tempSecond[i] = data[i+j]; // fix
return tempSecond;
}
The wiki article has a somewhat optimized approach for top down merge sort:
https://en.wikipedia.org/wiki/Merge_sort#Top-down_implementation

The problems in your code come from the confusing convention to specify ranges with included boundaries. You should instead consider the upper bound to be excluded: this would avoid the numerous +1/-1 adjustments required for the included convention, some of which are inconsistent in your code:
leftHalf creates an array with length (data.length / 2) + 1, including the element at offset mid = data.length / 2. This is what the merge method expects, but is not optimal as it would make unbalanced halves for even sized arrays, and more problematically would return a 2 element slice for a 2 element array, which causes an infinite recursion and explains the Stack Overflow exception you get.
rightHalf also creates an array with length (data.length / 2) + 1, which is incorrect, the length should be array with lengthdata.length - ((data.length / 2) + 1)`, which would be an empty array for a 2 element array.
Furthermore, rightHalf uses two nested for loops to copy the values from the argument array, which is incorrect.
your index range into the right array is incorrect in the merge method.
the index into data is incorrect in data[leftPos + mergePos] = mergedNumbers[mergePos]; it should be just:
data[mergePos] = mergedNumbers[mergePos];
Here is a modified version, with a less error prone convention:
// sort elements in data in place
public static void mergeSort(int[] data) {
if (data.length > 1) {
int[] left = firstHalf(data);
int[] right = secondHalf(data);
mergeSort(left);
mergeSort(right);
merge(data, left, right);
}
}
public static void merge(int[] data, int[] left, int[] right) {
int leftLength = left.length;
int rightLength = right.length;
int length = leftLength + rightLength;
int mergedNumbers[] = new int[length]; //Temp array to received the merged array
int leftPos = 0;
int rightPos = 0;
int mergePos = 0;
while (leftPos < leftLength && rightPos < rightLength) {
if (left[leftPos] <= right[rightPos]) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
} else {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
}
// copy the remaining entries in the left half
while (leftPos < leftLength) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
}
// copy the remaining entries in the right half
while (rightPos < rightLength) {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
// copy the values back to the original array
for (mergePos = 0; mergePos < length; mergePos++) {
data[mergePos] = mergedNumbers[mergePos];
}
}
public static int[] firstHalf(int[] data) {
int leftLength = data.length / 2;
int[] tempFirst = new int[leftLength];
for (int i = 0; i < leftLength; i++) {
tempFirst[i] = data[i];
}
return tempFirst;
}
public static int[] secondHalf(int[] data) {
int leftLength = data.length / 2;
int rightLength = data.length - leftLength;
int[] tempSecond = new int[rightLength];
for (int i = 0; i < rightLength; i++) {
tempSecond[i] = data[LeftLength + i];
}
return tempSecond;
}

Find the max value of the same length nails after hammered

I'm trying to solve this problem:
Given an array of positive integers, and an integer Y, you are allowed to replace at most Y array-elements with lesser values. Your goal is for the array to end up with as large a subset of identical values as possible. Return the size of this largest subset.
The array is originally sorted in increasing order, but you do not need to preserve that property.
So, for example, if the array is [10,20,20,30,30,30,40,40,40] and Y = 3, the result should be 6, because you can get six 30s by replacing the three 40s with 30s. If the array is [20,20,20,40,50,50,50,50] and Y = 2, the result should be 5, because you can get five 20s by replacing two of the 50s with 20s.
Below is my solution with O(nlogn) time complexity. (is that right?) I wonder if I can further optimize this solution?
Thanks in advance.
public class Nails {
public static int Solutions(int[] A, int Y) {
int N = A.length;
TreeMap < Integer, Integer > nailMap = new TreeMap < Integer, Integer > (Collections.reverseOrder());
for (int i = 0; i < N; i++) {
if (!nailMap.containsKey(A[i])) {
nailMap.put(A[i], 1);
} else {
nailMap.put(A[i], nailMap.get(A[i]) + 1);
}
}
List < Integer > nums = nailMap.values().stream().collect(Collectors.toList());
if (nums.size() == 1) {
return nums.get(0);
}
//else
int max = nums.get(0);
int longer = 0;
for (int j = 0; j < nums.size(); j++) {
int count = 0;
if (Y < longer) {
count = Y + nums.get(j);
} else {
count = longer + nums.get(j);
}
if (max < count) {
max = count;
}
longer += nums.get(j);
}
return max;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
String[] input = scanner.nextLine().replaceAll("\\[|\\]", "").split(",");
System.out.println(Arrays.toString(input));
int[] A = new int[input.length - 1];
int Y = Integer.parseInt(input[input.length - 1]);
for (int i = 0; i < input.length; i++) {
if (i < input.length - 1) {
A[i] = Integer.parseInt(input[i]);
} else {
break;
}
}
int result = Solutions(A, Y);
System.out.println(result);
}
}
}

A C++ implementation would like the following where A is the sorted pin size array and K is the number of times the pins can be hammered.
{1,1,3,3,4,4,4,5,5}, K=2 should give 5 as the answer
{1,1,3,3,4,4,4,5,5,6,6,6,6,6,6}, K=2 should give 6 as the answer
int maxCount(vector<int>& A, int K) {
int n = A.size();
int best = 0;
int count = 1;
for (int i = 0; i < n-K-1; i++) {
if (A[i] == A[i + 1])
count = count + 1;
else
count = 1;
if (count > best)
best = count;
}
int result = max(best+K, min(K+1, n));
return result;
}

Since the array is sorted to begin with, a reasonably straightforward O(n) solution is, for each distinct value, to count how many elements have that value (by iteration) and how many elements have a greater value (by subtraction).
public static int doIt(final int[] array, final int y) {
int best = 0;
int start = 0;
while (start < array.length) {
int end = start;
while (end < array.length && array[end] == array[start]) {
++end;
}
// array[start .. (end-1)] is now the subarray consisting of a
// single value repeated (end-start) times.
best = Math.max(best, end - start + Math.min(y, array.length - end));
start = end; // skip to the next distinct value
}
assert best >= Math.min(y + 1, array.length); // sanity-check
return best;
}

First, iterate through all the nails and create a hash H that stores the number of nails for each size. For [1,2,2,3,3,3,4,4,4], H should be:
size count
1 : 1
2 : 2
3 : 3
4 : 3
Now create an little algorithm to evaluate the maximum sum for each size S, given Y:
BestForSize(S, Y){
total = H[S]
while(Y > 0){
S++
if(Y >= H[S] and S < biggestNailSize){
total += H[S]
Y -= H[S]
}
else{
total += Y
Y = 0
}
}
return total;
}
Your answer should be max(BestForSize(0, Y), BestForSize(1, Y), ..., BestForSize(maxSizeOfNail, Y)).
The complexity is O(n²). A tip to optimize is to start from the end. For example, after you have the maximum value of nails in the size 4, how can you use your answer to find the maximum number of size 3?

Here is my java implementation: First I build a reversed map of each integer and its occurence for example {1,1,1,1,3,3,4,4,5,5} would give {5=2, 4=2, 3=2, 1=4}, then for each integer I calculate the max occurence that we can get of it regarding the K and the occurences of the highest integers in the array.
public static int ourFunction(final int[] A, final int K) {
int length = A.length;
int a = 0;
int result = 0;
int b = 0;
int previousValue = 0;
TreeMap < Integer, Integer > ourMap = new TreeMap < Integer, Integer > (Collections.reverseOrder());
for (int i = 0; i < length; i++) {
if (!ourMap.containsKey(A[i])) {
ourMap.put(A[i], 1);
} else {
ourMap.put(A[i], ourMap.get(A[i]) + 1);
}
}
for (Map.Entry<Integer, Integer> entry : ourMap.entrySet()) {
if( a == 0) {
a++;
result = entry.getValue();
previousValue = entry.getValue();
} else {
if( K < previousValue)
b = K;
else
b = previousValue;
if ( b + entry.getValue() > result )
result = b + entry.getValue();
previousValue += entry.getValue();
}
}
return result;
}

Since the array is sorted, we can have an O(n) solution by iterating and checking if current element is equals to previous element and keeping track of the max length.
static int findMax(int []a,int y) {
int n = a.length,current = 1,max = 0,diff = 0;
for(int i = 1; i< n; i++) {
if(a[i] == a[i-1]) {
current++;
diff = Math.min(y, n-i-1);
max = Math.max(max, current+diff);
}else {
current = 1;
}
}
return max;
}

given int array is not sorted than you should sort
public static int findMax(int []A,int K) {
int current = 1,max = 0,diff = 0;
List<Integer> sorted=Arrays.stream(A).sorted().boxed().collect(Collectors.toList());
for(int i = 1; i< sorted.size(); i++) {
if(sorted.get(i).equals(sorted.get(i-1))) {
current++;
diff = Math.min(K, sorted.size()-i-1);
max = Math.max(max, current+diff);
}else {
current = 1;
}
}
return max;
}
public static void main(String args[]) {
List<Integer> A = Arrays.asList(3,1,5,3,4,4,3,3,5,5,5,1);
int[] Al = A.stream().mapToInt(Integer::intValue).toArray();
int result=findMax(Al, 5);
System.out.println(result);
}

Automatic increment array java

Hello I have got this basically fully working sorted vector , the problem here is however that I can only initialize the array to a fixed size before inserting any values , so for example I can initialize 5 but if I want to insert 6 items it gives me a null pointer exception .
I think I do understand what is happening however I would like anybody to show me a solution how the array size can be increased automatically every time I want to insert something .
( Without having to use any inbuilt java functionalities like ArrayList )
Thank you
package ads2;
public class SortedVector2
{
private int length;
private int maximum;
private int growby;
private int temp;
private int x = 0;
private int high;
private int middle;
private int low;
String[] data;
public SortedVector2()
{
length = 0;
maximum = 5;
data = new String[maximum];
}
public void AddItem(String value)
{
/*if (length == maximum)
{
maximum += 200000;
*/
data[length] = value;
length++;
// SetSorted();
// SetSorted(data);
}
public void SetSorted()
{
for (int j = 0; j < data.length - 1; j++) {
if (data[j].compareTo(data[j + 1]) > -1) {
String temp = data[j];
data[j] = data[j + 1];
data[j + 1] = temp;
}
}
for (String s : data) {
System.out.println(s);
}
// private String[] data;
/*
for(int i = data.length-1; i >= 0; i--) {
for(int j = 0; j < i; j++) {
if(data[j].compareTo(data[j + 1]) > -1) {
String temp = data[j];
data[j] = data[j + 1];
data[j + 1] = temp;
}
}
} for (String s : data) {
System.out.println(s);
}
*/
}
public void SetGrowBy(int growby)
{
maximum += growby;
}
public int GetCapacity()
{
return maximum;
}
public int GetNoOfItems()
{
return length;
}
public String GetItemByIndex(int index)
{
return data[index];
}
public int FindItem(String search)
{
for (x=0;x<=length; )
{
middle =((low + high)/2);
if (data[middle].compareTo(search)==0)
{
return middle;
}
else if (data[middle].compareTo(search)<0)
{
low = middle;
x++;
return FindItem(search);
}
else
{
high = middle;
x++;
return FindItem(search);
}
}
return -1;
}
public boolean Exists(String search)
{
boolean output;
int y;
y = 0;
while (data[y] != search && (length - 1) > y)
{
++y;
}
if (data[y] == search)
{
output = true;
} else
{
output = false;
}
y = 0;
return output;
}
public void InsertItem(int index, String value)
{
if (length == maximum)
{
maximum += 200000;
}
for(int i = length - 1; i >= index; --i)
{
data[i + 1] = data[i];
}
data[index] = value;
length++;
}
public void DeleteItem(int index)
{
for(int x = index; x < length - 2; ++x)
{
data[x] = data[x + 1];
}
length--;
}
public String toString()
{
String res = "";
for (int i=0; i<length; i++)
res+=data[i] + "; ";
return res;
}
}

You have to do what the implementers of ArrayList did. When you try to add an element when the array is full, you create a larger array, copy the existing elements to it and add the new element.

To increase array size dynamically use Collection framework interface List,
It has implementation ArrayList,Vector and LinkedList use any one in them.
Or, Simply create copyArray(String[]) api which will give you array with increased capacity.
public String[] copyArray(String[] oldArray){
int capacity = oldArray.length * 2;
return Arrays.copyOf(oldArray, capacity);
}
String[] data = copyArray(data) // pass array

I think you've got all the basic variables you need to do what you need to do: just check if the size equals the capacity when you are adding an item and if it does reallocate the array:
if (size == capacity) {
capacity += growby;
data = Arrays.copyOf(data, capacity);
}
That's pretty much all ArrayList does.

You need to re-allocate when increasing the size of the data buffer, for example,
public void InsertItem(int index, String value)
{
String[] data2;
if (length == (maximum-1))
{
maximum += 5; // increment size in lot of 5
data2 = new String[maximum);
for (int ii = 0; ii < length; ii++)
{
data2[ii] = date[ii];
}
data = data2; // re-assign with increased size
}
for(int i = length - 1; i >= index; --i)
{
data[i + 1] = data[i];
}
data[index] = value;
length++;
}

In software engineering there is a saying, "Don't reinvent the wheel" - which emphasizes us on using the existing archetype. Because they are tested and used by for long period of time. So it's better to use ArrayList for regular/professional purpose.
But it if it is for learning purpose then you can chose any one from the previous answers.

Implementing a binary insertion sort using binary search in Java

I'm having trouble combining these two algorithms together. I've been asked to modify Binary Search to return the index that an element should be inserted into an array. I've been then asked to implement a Binary Insertion Sort that uses my Binary Search to sort an array of randomly generated ints.
My Binary Search works the way it's supposed to, returning the correct index whenever I test it alone. I wrote out Binary Insertion Sort to get a feel for how it works, and got that to work as well. As soon as I combine the two together, it breaks. I know I'm implementing them incorrectly together, but I'm not sure where my problem lays.
Here's what I've got:
public class Assignment3
{
public static void main(String[] args)
{
int[] binary = { 1, 7, 4, 9, 10, 2, 6, 12, 3, 8, 5 };
ModifiedBinaryInsertionSort(binary);
}
static int ModifiedBinarySearch(int[] theArray, int theElement)
{
int leftIndex = 0;
int rightIndex = theArray.length - 1;
int middleIndex = 0;
while(leftIndex <= rightIndex)
{
middleIndex = (leftIndex + rightIndex) / 2;
if (theElement == theArray[middleIndex])
return middleIndex;
else if (theElement < theArray[middleIndex])
rightIndex = middleIndex - 1;
else
leftIndex = middleIndex + 1;
}
return middleIndex - 1;
}
static void ModifiedBinaryInsertionSort(int[] theArray)
{
int i = 0;
int[] returnArray = new int[theArray.length + 1];
for(i = 0; i < theArray.length; i++)
{
returnArray[ModifiedBinarySearch(theArray, theArray[i])] = theArray[i];
}
for(i = 0; i < theArray.length; i++)
{
System.out.print(returnArray[i] + " ");
}
}
}
The return value I get for this when I run it is 1 0 0 0 0 2 0 0 3 5 12. Any suggestions?
UPDATE: updated ModifiedBinaryInsertionSort
static void ModifiedBinaryInsertionSort(int[] theArray)
{
int index = 0;
int element = 0;
int[] returnArray = new int[theArray.length];
for (int i = 1; i < theArray.lenght - 1; i++)
{
element = theArray[i];
index = ModifiedBinarySearch(theArray, 0, i, element);
returnArray[i] = element;
while (index >= 0 && theArray[index] > element)
{
theArray[index + 1] = theArray[index];
index = index - 1;
}
returnArray[index + 1] = element;
}
}

Here is my method to sort an array of integers using binary search.
It modifies the array that is passed as argument.
public static void binaryInsertionSort(int[] a) {
if (a.length < 2)
return;
for (int i = 1; i < a.length; i++) {
int lowIndex = 0;
int highIndex = i;
int b = a[i];
//while loop for binary search
while(lowIndex < highIndex) {
int middle = lowIndex + (highIndex - lowIndex)/2; //avoid int overflow
if (b >= a[middle]) {
lowIndex = middle+1;
}
else {
highIndex = middle;
}
}
//replace elements of array
System.arraycopy(a, lowIndex, a, lowIndex+1, i-lowIndex);
a[lowIndex] = b;
}
}

How an insertion sort works is, it creates a new empty array B and, for each element in the unsorted array A, it binary searches into the section of B that has been built so far (From left to right), shifts all elements to the right of the location in B it choose one right and inserts the element in. So you are building up an at-all-times sorted array in B until it is the full size of B and contains everything in A.
Two things:
One, the binary search should be able to take an int startOfArray and an int endOfArray, and it will only binary search between those two points. This allows you to make it consider only the part of array B that is actually the sorted array.
Two, before inserting, you must move all elements one to the right before inserting into the gap you've made.

I realize this is old, but the answer to the question is that, perhaps a little unintuitively, "Middleindex - 1" will not be your insertion index in all cases.
If you run through a few cases on paper the problem should become apparent.
I have an extension method that solves this problem. To apply it to your situation, you would iterate through the existing list, inserting into an empty starting list.
public static void BinaryInsert<TItem, TKey>(this IList<TItem> list, TItem item, Func<TItem, TKey> sortfFunc)
where TKey : IComparable
{
if (list == null)
throw new ArgumentNullException("list");
int min = 0;
int max = list.Count - 1;
int index = 0;
TKey insertKey = sortfFunc(item);
while (min <= max)
{
index = (max + min) >> 1;
TItem value = list[index];
TKey compKey = sortfFunc(value);
int result = compKey.CompareTo(insertKey);
if (result == 0)
break;
if (result > 0)
max = index - 1;
else
min = index + 1;
}
if (index <= 0)
index = 0;
else if (index >= list.Count)
index = list.Count;
else
if (sortfFunc(list[index]).CompareTo(insertKey) < 0)
++index;
list.Insert(index, item);
}

Dude, I think you have some serious problem with your code. Unfortunately, you are missing the fruit (logic) of this algorithm. Your divine goal here is to get the index first, insertion is a cake walk, but index needs some sweat. Please don't see this algorithm unless you gave your best and desperate for it. Never give up, you already know the logic, your goal is to find it in you. Please let me know for any mistakes, discrepancies etc. Happy coding!!
public class Insertion {
private int[] a;
int n;
int c;
public Insertion()
{
a = new int[10];
n=0;
}
int find(int key)
{
int lowerbound = 0;
int upperbound = n-1;
while(true)
{
c = (lowerbound + upperbound)/2;
if(n==0)
return 0;
if(lowerbound>=upperbound)
{
if(a[c]<key)
return c++;
else
return c;
}
if(a[c]>key && a[c-1]<key)
return c;
else if (a[c]<key && a[c+1]>key)
return c++;
else
{
if(a[c]>key)
upperbound = c-1;
else
lowerbound = c+1;
}
}
}
void insert(int key)
{
find(key);
for(int k=n;k>c;k--)
{
a[k]=a[k-1];
}
a[c]=key;
n++;
}
void display()
{
for(int i=0;i<10;i++)
{
System.out.println(a[i]);
}
}
public static void main(String[] args)
{
Insertion i=new Insertion();
i.insert(56);
i.insert(1);
i.insert(78);
i.insert(3);
i.insert(4);
i.insert(200);
i.insert(6);
i.insert(7);
i.insert(1000);
i.insert(9);
i.display();
}
}