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

The minimum value of x is x = ceil((diff -s+ (n-i)* a) / (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 so assign x =0.
• Else find the value of x which is the value to be decremented in arr 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<``int``> 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 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;` `        ``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<``int``> 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 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 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 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;` `      ``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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