Given an array and a value k. We have to find the maximum number of equal elements possible for the array so that we can increase the elements of the array by incrementing a total of at-most k.

Examples:

Input : array = { 2, 4, 9 }, k = 3
Output : 2
We are allowed to do at most three increments. We can make two elements 4 by increasing 2 by 2. Note that we can not make two elements 9 as converting 4 to 9 requires 5 increments.

Input : array = { 5, 5, 3, 1 }, k = 5
Output : 3
Explanation: Here 1st and 2nd elements are equal. Then we can increase 3rd element 3 upto 5. Then k becomes (k-2) = 3. Now we can’t increase 1 to 5 because k value is 3 and we need 4 for the updation. Thus equal elements possible are 3. Here we can also increase 1 to 5. Then also we have 3 because we can’t update 3 to 5.

Input : array = { 5, 5, 3, 1 }, k = 6
Output : 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: In the naive approach we have an algorithm in O(n^2) time in which we check for each element how many other elements can be incremented so that they will become equal to them.

Efficient Approach: In this approach, first we will sort the array. Then we maintain two arrays. First is prefix sum array which stores the prefix sum of the array and another is maxx[] array which stores the maximum element found till every point, i.e., max[i] means maximum element from 1 to i. After storing these values in prefix[] array and maxx[] array, we do the binary search from 1 to n(number of elements of the array) to calculate how many elements which can be incremented to make them equal. In the binary search, we use one function in which we determine what is the number of elements can be incremented to make them equal to a single value.

## C++

 // C++ program to find maximum elements that can  // be made equal with k updates #include using namespace std;    // Function to calculate the maximum number of // equal elements possible with atmost K increment // of values .Here we have done sliding window // to determine that whether there are x number of // elements present which on increment will become // equal. The loop here will run in fashion like // 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1 bool ElementsCalculationFunc(int pre[], int maxx[],                                int x, int k, int n) {     for (int i = 0, j = x; j <= n; j++, i++) {            // It can be explained with the reasoning         // that if for some x number of elements         // we can update the values then the         // increment to the segment (i to j having         // length -> x) so that all will be equal is         // (x*maxx[j]) this is the total sum of         // segment and (pre[j]-pre[i]) is present sum         // So difference of them should be less than k         // if yes, then that segment length(x) can be          // possible return true         if (x * maxx[j] - (pre[j] - pre[i]) <= k)             return true;     }     return false; }    void MaxNumberOfElements(int a[], int n, int k) {     // sort the array in ascending order     sort(a, a + n);     int pre[n + 1]; // prefix sum array     int maxx[n + 1]; // maximum value array        // Initializing the prefix array     // and maximum array     for (int i = 0; i <= n; ++i) {         pre[i] = 0;         maxx[i] = 0;     }        // set the first element of both     // array     maxx[0] = a[0];     pre[0] = a[0];     for (int i = 1; i < n; i++) {            // Calculating prefix sum of the array         pre[i] = pre[i - 1] + a[i];            // Calculating max value upto that position         // in the array         maxx[i] = max(maxx[i - 1], a[i]);     }        // Binary seach applied for     // computation here     int l = 1, r = n, ans;     while (l < r) {         int mid = (l + r) / 2;            if (ElementsCalculationFunc(pre, maxx,                               mid - 1, k, n)) {             ans = mid;             l = mid + 1;         }         else             r = mid - 1;     }        // printing result     cout << ans << "\n"; }    int main() {     int arr[] = { 2, 4, 9 };     int n = sizeof(arr) / sizeof(arr[0]);     int k = 3;     MaxNumberOfElements(arr, n, k);     return 0; }

## Java

 // java program to find maximum elements that can  // be made equal with k updates    import java.util.Arrays; public class GFG {            // Function to calculate the maximum number of     // equal elements possible with atmost K increment     // of values .Here we have done sliding window     // to determine that whether there are x number of     // elements present which on increment will become     // equal. The loop here will run in fashion like     // 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1     static boolean ElementsCalculationFunc(int pre[],                        int maxx[], int x, int k, int n)     {         for (int i = 0, j = x; j <= n; j++, i++) {                     // It can be explained with the reasoning             // that if for some x number of elements             // we can update the values then the             // increment to the segment (i to j having             // length -> x) so that all will be equal is             // (x*maxx[j]) this is the total sum of             // segment and (pre[j]-pre[i]) is present sum             // So difference of them should be less than k             // if yes, then that segment length(x) can be              // possible return true             if (x * maxx[j] - (pre[j] - pre[i]) <= k)                 return true;         }         return false;     }             static void MaxNumberOfElements(int a[], int n, int k)     {         // sort the array in ascending order         Arrays.sort(a);         int []pre = new int[n + 1]; // prefix sum array         int []maxx = new int[n + 1]; // maximum value array                 // Initializing the prefix array         // and maximum array         for (int i = 0; i <= n; ++i) {             pre[i] = 0;             maxx[i] = 0;         }                 // set the first element of both         // array         maxx[0] = a[0];         pre[0] = a[0];         for (int i = 1; i < n; i++) {                     // Calculating prefix sum of the array             pre[i] = pre[i - 1] + a[i];                     // Calculating max value upto that position             // in the array             maxx[i] = Math.max(maxx[i - 1], a[i]);         }                 // Binary seach applied for         // computation here         int l = 1, r = n, ans=0;         while (l < r) {                             int mid = (l + r) / 2;                     if (ElementsCalculationFunc(pre, maxx,                                    mid - 1, k, n))             {                 ans = mid;                 l = mid + 1;             }             else                 r = mid - 1;         }                 // printing result         System.out.print((int)ans + "\n");     }              public static void main(String args[]) {                    int arr[] = { 2, 4, 9 };         int n = arr.length;         int k = 3;                     MaxNumberOfElements(arr, n, k);                 } }    // This code is contributed by Sam007

## Python3

 # Python3 program to find maximum elements  # that can be made equal with k updates    # Function to calculate the maximum number of # equal elements possible with atmost K increment # of values .Here we have done sliding window # to determine that whether there are x number of # elements present which on increment will become # equal. The loop here will run in fashion like # 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1 def ElementsCalculationFunc(pre, maxx,                             x, k, n):        i = 0     j = x     while j <= n:            # It can be explained with the reasoning         # that if for some x number of elements         # we can update the values then the         # increment to the segment (i to j having         # length -> x) so that all will be equal is         # (x*maxx[j]) this is the total sum of         # segment and (pre[j]-pre[i]) is present sum         # So difference of them should be less than k         # if yes, then that segment length(x) can be          # possible return true         if (x * maxx[j] - (pre[j] - pre[i]) <= k):             return True            i += 1         j += 1            return False    def MaxNumberOfElements( a, n, k):        # sort the array in ascending order     a.sort()     pre = [0] * (n + 1) # prefix sum array     maxx = [0] * (n + 1) # maximum value array        # Initializing the prefix array     # and maximum array     for i in range (n + 1):         pre[i] = 0         maxx[i] = 0        # set the first element of both     # array     maxx[0] = a[0]     pre[0] = a[0]     for i in range(1, n):            # Calculating prefix sum of the array         pre[i] = pre[i - 1] + a[i]            # Calculating max value upto that         # position in the array         maxx[i] = max(maxx[i - 1], a[i])        # Binary seach applied for     # computation here     l = 1     r = n     while (l < r) :         mid = (l + r) // 2            if (ElementsCalculationFunc(pre, maxx,                                     mid - 1, k, n)):             ans = mid             l = mid + 1                    else:             r = mid - 1        # printing result     print (ans)    # Driver Code if __name__ == "__main__":        arr = [2, 4, 9 ]     n = len(arr)     k = 3     MaxNumberOfElements(arr, n, k)    # This code is contributed by Ita_c

## C#

 // C# program to find maximum elements that can  // be made equal with k updates using System;    class GFG {            // Function to calculate the maximum number of     // equal elements possible with atmost K increment     // of values .Here we have done sliding window     // to determine that whether there are x number of     // elements present which on increment will become     // equal. The loop here will run in fashion like     // 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1     static bool ElementsCalculationFunc(int []pre,                        int []maxx, int x, int k, int n)     {         for (int i = 0, j = x; j <= n; j++, i++) {                    // It can be explained with the reasoning             // that if for some x number of elements             // we can update the values then the             // increment to the segment (i to j having             // length -> x) so that all will be equal is             // (x*maxx[j]) this is the total sum of             // segment and (pre[j]-pre[i]) is present sum             // So difference of them should be less than k             // if yes, then that segment length(x) can be              // possible return true             if (x * maxx[j] - (pre[j] - pre[i]) <= k)                 return true;         }         return false;     }            static void MaxNumberOfElements(int []a, int n, int k)     {         // sort the array in ascending order         Array.Sort(a);         int []pre = new int[n + 1]; // prefix sum array         int []maxx = new int[n + 1]; // maximum value array                // Initializing the prefix array         // and maximum array         for (int i = 0; i <= n; ++i) {             pre[i] = 0;             maxx[i] = 0;         }                // set the first element of both         // array         maxx[0] = a[0];         pre[0] = a[0];         for (int i = 1; i < n; i++) {                    // Calculating prefix sum of the array             pre[i] = pre[i - 1] + a[i];                    // Calculating max value upto that position             // in the array             maxx[i] = Math.Max(maxx[i - 1], a[i]);         }                // Binary seach applied for         // computation here         int l = 1, r = n, ans=0;         while (l < r) {                            int mid = (l + r) / 2;                    if (ElementsCalculationFunc(pre, maxx,                                    mid - 1, k, n))             {                 ans = mid;                 l = mid + 1;             }             else                 r = mid - 1;         }                // printing result         Console.Write ((int)ans + "\n");     }             // Driver code     public static void Main()     {         int []arr = { 2, 4, 9 };         int n = arr.Length;         int k = 3;                    MaxNumberOfElements(arr, n, k);     } }    // This code is contributed by Sam007

## PHP

 x) so that all will be equal is         // (x*maxx[j]) this is the total sum of         // segment and (pre[j]-pre[i]) is present sum         // So difference of them should be less than k         // if yes, then that segment length(x) can be          // possible return true         if (\$x * \$maxx[\$j] - (\$pre[\$j] - \$pre[\$i]) <= \$k)             return true;     }     return false; }    function  MaxNumberOfElements(\$a, \$n, \$k) {     // sort the array in ascending order     sort(\$a);     \$pre[\$n + 1]=array(); // prefix sum array     \$maxx[\$n + 1]=array(); // maximum value array        // Initializing the prefix array     // and maximum array     for (\$i = 0; \$i <= \$n; ++\$i) {         \$pre[\$i] = 0;         \$maxx[\$i] = 0;     }        // set the first element of both     // array     \$maxx[0] = \$a[0];     \$pre[0] = \$a[0];     for (\$i = 1; \$i < \$n; \$i++) {            // Calculating prefix sum of the array         \$pre[\$i] = \$pre[\$i - 1] + \$a[\$i];            // Calculating max value upto that position         // in the array         \$maxx[\$i] = max(\$maxx[\$i - 1], \$a[\$i]);     }        // Binary seach applied for     // computation here     \$l = 1;     \$r = \$n;     \$ans;     while (\$l < \$r) {         \$mid = (\$l + \$r) / 2;            if (ElementsCalculationFunc(\$pre, \$maxx,                             \$mid - 1, \$k, \$n)) {             \$ans = \$mid;             \$l = \$mid + 1;         }         else             \$r = \$mid - 1;     }        // printing result     echo  \$ans , "\n"; } //Code driven     \$arr = array(2, 4, 9 );     \$n = sizeof(\$arr) / sizeof(\$arr[0]);     \$k = 3;     MaxNumberOfElements(\$arr, \$n, \$k);       #This code is contributed by akt_mit. ?>

Output:

2

Time Complexity :O(nlog(n))
Space Complexity : O(n)

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