# Minimum operations to make Array sum at most S from given Array

• Last Updated : 07 Jan, 2022

Given an array arr[], of size N and an integer S, the task is to find the minimum operations to make the sum of the array less than or equal to S. In each operation:

• Any element can be chosen and can be decremented by 1, or
• Can be replaced by any other element in the array.

Examples:

Input: arr[]= {1, 2, 1 ,3, 1 ,2, 1}, S= 8
Output: 2
Explanation: Initially sum of the array is 11.
Now decrease 1 at index 0 to 0 and Replace 3 by 0.
The sum becomes  7 < 8. So 2 operations.

Input: arr[] = {1,2,3,4}, S= 11
Output: 0
Explanation: Sum is already < =11 so 0 operations.

Approach: This problem can be solved using the greedy approach and suffix sum by sorting the array. Applying the 1st operation on the minimum element any number of times and then applying the 2nd operation on the suffixes by replacing it with the minimum element after the first operation gives the minimum operations.

First sort the array. Consider performing x operations of 1st type on the arr[0] and then performing the 2nd operation on the suffix of the array of length i. Also consider the sum for this suffix of length i is sufSum.

Sum of the modified array must be <=S
So, the difference to be  subtracted from the sum must be (diff)>= sum – S.

If x operations of type 1 is done on minimum element and type 2 operations are done the suffix of the array from [i,n) the sum of the decreased array is

cost = x + s – (n-i) * (a[0] – x)
cost = (n-i+1)* x-(n-i)* a[0]  +s
cost >= sum – S = diff
s – (n-i) * a[0] + (n-i+1) *x >= diff
so x >= (diff – s+(n-i)* a[0]) / (n-i+1)

The minimum value of x is x = ceil((diff -s+ (n-i)* a[0]) / (n-i+1))

So the total operations are x (type-1) + (n-i) type-2

Follow these steps to solve the above problems:

• Initialize a variable sum = 0 and the size of the array to N.
• Iterate through the vector and find the sum of the array.
• If sum < = S print 0 and return.
• Sort the vector and assign diff = sum-S.
• Initialize ops = sum-S which is the maximum possible operations.
• Initialize s =0 which stores the suffix sum of the vector.
• Now traverse from the end of the vector using for loop.
• Keep track of suffix sum in s variable.
• Initialize a dec variable which is the value to be decremented from the suffix of the array
• If s-dec is greater than or equal to diff there is no need to decrement arr[0] so assign x =0.
• Else find the value of x which is the value to be decremented in arr[0] and find the minimum operations.
• Print the minimum operations

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;   // Function to divide and get the ceil value int ceil_div(int a, int b) {     return a / b + ((a ^ b) > 0 && a % b); }   // Function to find the minimum cost void minimum_cost(vector arr, int S) {     int sum = 0;     int n = arr.size();           // Find the sum of the array     for (int i = 0; i < arr.size(); i++) {         sum += arr[i];     }           // If sum <= S no operations required     if (sum <= S) {         cout << 0 << endl;         return;     }           // Sort the array     sort(arr.begin(), arr.end());       int diff = sum - S;           // Maximum it requires sum-S operations     // by decrementing     // the arr[0] by 1     int ops = sum - S;     // suffix sum     int s = 0;     int x;       for (int i = n - 1; i > 0; i--) {         s += arr[i];                   // If replacing the last elements         // with doing the first operation         // x = 0 Decrementing the a[i] from         // the suffix [i,n-1]         int dec = (n - i) * arr[0];         if (s - dec >= diff) {             x = 0;         }                   // Find how times the first element         // should be decremented by 1 and         // incremented by 1 which is x         else {             x = max(ceil_div((diff - s + dec),                              (n - i + 1)), 0);         }                   // First operation + second operation         if (x + n - i < ops) {             ops = x + n - i;         }     }       // Print the operations     cout << ops << endl; }   // Driver code int main() {     // Initialize the array     vector arr = { 1, 2, 1, 3, 1, 2, 1 };     int S = 8;           // Function call     minimum_cost(arr, S);       return 0; }

## Java

 // Java program for the above approach   import java.util.Arrays;   class GFG {     // Function to divide and get the ceil value   static int ceil_div(int a, int b) {     int temp = 0;     if (((a ^ b) > 0) && ((a % b) > 0)) {       temp = 1;     }     return (a / b) + temp;   }     // Function to find the minimum cost   static void minimum_cost(int[] arr, int S) {     int sum = 0;     int n = arr.length;       // Find the sum of the array     for (int i = 0; i < arr.length; i++) {       sum += arr[i];     }       // If sum <= S no operations required     if (sum <= S) {       System.out.println(0);       return;     }       // Sort the array     Arrays.sort(arr);       int diff = sum - S;       // Maximum it requires sum-S operations     // by decrementing     // the arr[0] by 1     int ops = sum - S;     // suffix sum     int s = 0;     int x;       for (int i = n - 1; i > 0; i--) {       s += arr[i];         // If replacing the last elements       // with doing the first operation       // x = 0 Decrementing the a[i] from       // the suffix [i,n-1]       int dec = (n - i) * arr[0];       if (s - dec >= diff) {         x = 0;       }         // Find how times the first element       // should be decremented by 1 and       // incremented by 1 which is x       else {         x = Math.max(ceil_div((diff - s + dec),                               (n - i + 1)), 0);       }         // First operation + second operation       if (x + n - i < ops) {         ops = x + n - i;       }     }       // Print the operations     System.out.println(ops);   }     // Driver code   public static void main(String args[])   {       // Initialize the array     int[] arr = { 1, 2, 1, 3, 1, 2, 1 };     int S = 8;       // Function call     minimum_cost(arr, S);   } }   // This code is contributed by saurabh_jaiswal.

## Python3

 # Python 3 program for the above approach   # Function to divide and get the ceil value def ceil_div(a,  b):       return a // b + ((a ^ b) > 0 and a % b)   # Function to find the minimum cost def minimum_cost(arr, S):       sum = 0     n = len(arr)       # Find the sum of the array     for i in range(len(arr)):         sum += arr[i]       # If sum <= S no operations required     if (sum <= S):         print(0)         return       # Sort the array     arr.sort()       diff = sum - S       # Maximum it requires sum-S operations     # by decrementing     # the arr[0] by 1     ops = sum - S     # suffix sum     s = 0       for i in range(n - 1, -1, -1):         s += arr[i]           # If replacing the last elements         # with doing the first operation         # x = 0 Decrementing the a[i] from         # the suffix [i,n-1]         dec = (n - i) * arr[0]         if (s - dec >= diff):             x = 0           # Find how times the first element         # should be decremented by 1 and         # incremented by 1 which is x         else:             x = max(ceil_div((diff - s + dec),                              (n - i + 1)), 0)           # First operation + second operation         if (x + n - i < ops):             ops = x + n - i       # Print the operations     print(ops)   # Driver code if __name__ == "__main__":       # Initialize the array     arr = [1, 2, 1, 3, 1, 2, 1]     S = 8       # Function call     minimum_cost(arr, S)       # This code is contributed by ukasp.

## C#

 // C# program for the above approach using System; class GFG {   // Function to divide and get the ceil value   static int ceil_div(int a, int b) {     int temp = 0;     if (((a ^ b) > 0) && ((a % b) > 0)) {       temp = 1;     }     return (a / b) + temp;   }     // Function to find the minimum cost   static void minimum_cost(int[] arr, int S) {     int sum = 0;     int n = arr.Length;       // Find the sum of the array     for (int i = 0; i < arr.Length; i++) {       sum += arr[i];     }       // If sum <= S no operations required     if (sum <= S) {       Console.WriteLine(0);       return;     }       // Sort the array     Array.Sort(arr);       int diff = sum - S;       // Maximum it requires sum-S operations     // by decrementing     // the arr[0] by 1     int ops = sum - S;     // suffix sum     int s = 0;     int x;       for (int i = n - 1; i > 0; i--) {       s += arr[i];         // If replacing the last elements       // with doing the first operation       // x = 0 Decrementing the a[i] from       // the suffix [i,n-1]       int dec = (n - i) * arr[0];       if (s - dec >= diff) {         x = 0;       }         // Find how times the first element       // should be decremented by 1 and       // incremented by 1 which is x       else {         x = Math.Max(ceil_div((diff - s + dec),                               (n - i + 1)), 0);       }         // First operation + second operation       if (x + n - i < ops) {         ops = x + n - i;       }     }       // Print the operations     Console.Write(ops);   }     // Driver code   public static void Main()   {       // Initialize the array     int[] arr = { 1, 2, 1, 3, 1, 2, 1 };     int S = 8;       // Function call     minimum_cost(arr, S);   } }   // This code is contributed by Samim Hossain Mondal.

## Javascript



Output

2

Time Complexity: O(N* logN)
Space Complexity: O(1)

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