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

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:

2

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