Reduce given Array to 0 by maximising sum of chosen elements
Last Updated :
23 Jul, 2025
Given an array arr[] containing N positive integers, the task is to maximize the array sum when after every sum operation all the remaining array elements decrease by 1.
Note: The value of an array element does not go below 0.
Examples:
Input: arr[] = {6, 2, 4, 5}
Output: 12
Explanation: Add 6 initially to the final sum.
The final sum becomes 6 and the remaining array elements {1, 3, 4}.
Add 3 with the sum. The sum becomes 9 and the remaining elements {0, 3}
Add 3 with the sum. The sum becomes 12 and only 0 remains in the array.
Add 0 with the sum. The sum remains unchanged.Input: arr[] = {5, 6, 4}
Output: 12
Naive approach: Find the maximum element from the array. Add it to the sum and decrease all other elements by 1 and change the current element to 0 so it is not repeated again in the loop. Do this process until all the elements become 0.
C++
// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
// Find maximum possible sum
int maxSum(int arr[], int N)
{
// Initialize ans with 0
int ans = 0;
// loop till atleast one element is greater than 0
while (1) {
// maximum element of array
int maxValueIndex = max_element(arr, arr + N) - arr;
// breaking condition when all elements become <=0
if (arr[maxValueIndex] <= 0)
break;
// adding value to answer
ans += arr[maxValueIndex];
arr[maxValueIndex] = 0;
// Iterate array
for (int i = 0; i < N; i++) {
arr[i]--;
}
}
return ans;
}
// Driver code
int main()
{
// Given array of values
int arr[] = { 6, 2, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << maxSum(arr, N);
return 0;
}
// Java code to implement the above approach
import java.util.*;
class GFG
{
// Find maximum possible sum
static int maxSum(int arr[], int N)
{
// Initialize ans with 0
int ans = 0;
// loop till atleast one element is greater than 0
while (true) {
// maximum element of array
int maxValue = Arrays.stream(arr).max().getAsInt();
;
int maxValueIndex = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == maxValue) {
maxValueIndex = i;
break;
}
}
// breaking condition when all elements become <=0
if (arr[maxValueIndex] <= 0)
break;
// adding value to answer
ans += arr[maxValueIndex];
arr[maxValueIndex] = 0;
// Iterate array
for (int i = 0; i < N; i++) {
arr[i]--;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
// Given array of values
int arr[] = { 6, 2, 4, 5 };
int N = arr.length;
// Function call
System.out.print(maxSum(arr, N));
}
}
// This code is contributed by gauravrajput1
# Python code to implement the above approach
# Find maximum possible sum
def maxSum(arr, N):
# Initialize ans with 0
ans = 0
# loop till atleast one element is greater than 0
while (1):
# maximum element of array
maxValueIndex = arr.index(max(arr))
# breaking condition when all elements become <=0
if (arr[maxValueIndex] <= 0):
break
# adding value to answer
ans += arr[maxValueIndex]
arr[maxValueIndex] = 0
# Iterate array
for i in range(0, N):
arr[i] -= 1
return ans
# Driver code
# Given array of values
arr = [6, 2, 4, 5]
N = len(arr)
# Function call
print(maxSum(arr, N))
# This code is contributed by ninja_hattori.
// C# code to implement the above approach
using System;
using System.Linq;
class GFG
{
// Find maximum possible sum
static int maxSum(int[] arr, int N)
{
// Initialize ans with 0
int ans = 0;
// loop till atleast one element is greater than 0
while (true) {
// maximum element of array
int maxValue = arr.Max();
int maxValueIndex = 0;
for (int i = 0; i < arr.Length; i++) {
if (arr[i] == maxValue) {
maxValueIndex = i;
break;
}
}
// breaking condition when all elements become <=0
if (arr[maxValueIndex] <= 0)
break;
// adding value to answer
ans += arr[maxValueIndex];
arr[maxValueIndex] = 0;
// Iterate array
for (int i = 0; i < N; i++) {
arr[i]--;
}
}
return ans;
}
// Driver code
public static int Main()
{
// Given array of values
int[] arr = { 6, 2, 4, 5 };
int N = arr.Length;
// Function call
Console.Write(maxSum(arr, N));
return 0;
}
}
// This code is contributed by Pushpesh Raj
<script>
// JavaScript code to implement the above approach
// Find maximum possible sum
function maxSum(arr, N)
{
// Initialize ans with 0
let ans = 0;
// loop till atleast one element is greater than 0
while (1) {
// maximum element of array
let maxValueIndex = arr.indexOf(Math.max(...arr));
// breaking condition when all elements become <=0
if (arr[maxValueIndex] <= 0)
break;
// adding value to answer
ans += arr[maxValueIndex];
arr[maxValueIndex] = 0;
// Iterate array
for (let i = 0; i < N; i++) {
arr[i]--;
}
}
return ans;
}
// Driver code
// Given array of values
let arr = [ 6, 2, 4, 5 ];
let N = arr.length;
// Function call
document.write(maxSum(arr, N));
// This code is contributed by shinjanpatra
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1), since no extra space has been taken.
Efficient Approach: The solution of the problem is based on the concept of sorting. As the values decrease in each step, the higher values should be added first with the final sum. Follow the steps mentioned below to solve the problem:
- Sort the array in descending order.
- Run a loop from 1 to N.
- For each index i the value added to the final sum will be (arr[i] - i) as it will be added after i decrement of the value.
- As the sum value cannot go below 0, add max(arr[i]-i, 0) to the final sum.
- Return the final sum.
Below is the implementation of the above approach.
C++
// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
// Find maximum possible sum
int maxSum(int arr[], int N)
{
// Initialize ans with 0
int ans = 0;
// Sort array in descending order
sort(arr, arr + N, greater<int>());
// Iterate array
for (int i = 0; i < N; i++) {
// Starting value
int value = arr[i];
// Actual value when being added
int current = max(0, value - i);
// Add actual value with ans
ans = ans + current;
}
return ans;
}
// Driver code
int main()
{
// Given array of values
int arr[] = { 6, 2, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << maxSum(arr, N);
return 0;
}
// Java code to implement the above approach
import java.util.*;
public class GFG
{
// Utility program to reverse an array
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swapping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
// Find maximum possible sum
static int maxSum(int[] arr, int N)
{
// Initialize ans with 0
int ans = 0;
// Sorting the array in ascending order
Arrays.sort(arr);
// Reversing the array
reverse(arr);
// Iterate array
for (int i = 0; i < N; i++) {
// Starting value
int value = arr[i];
// Actual value when being added
int current = Math.max(0, value - i);
// Add actual value with ans
ans = ans + current;
}
return ans;
}
// Driver code
public static void main(String args[])
{
// Given array of values
int[] arr = { 6, 2, 4, 5 };
int N = arr.length;
// Function call
System.out.println(maxSum(arr, N));
}
}
// This code is contributed by Samim Hossain Mondal.
# Python code to implement the above approach
# Find maximum possible sum
def maxSum(arr, N):
# Initialize ans with 0
ans = 0
# Sort array in descending order
arr.sort(reverse = True)
# Iterate array
for i in range(N):
#Starting value
value = arr[i]
# Actual value when being added
current = max(0, value - i)
# Add actual value with ans
ans = ans + current
return ans
# Driver code
if __name__ == "__main__":
# Given array of values
arr = [ 6, 2, 4, 5 ]
N = len(arr)
# Function call
print(maxSum(arr, N))
# This code is contributed by hrithikgarg03188.
// C# code to implement the above approach
using System;
class GFG
{
// Find maximum possible sum
static int maxSum(int[] arr, int N)
{
// Initialize ans with 0
int ans = 0;
// Sort array in descending order
Array.Sort<int>(
arr, delegate(int m, int n) { return n - m; });
// Iterate array
for (int i = 0; i < N; i++) {
// Starting value
int value = arr[i];
// Actual value when being added
int current = Math.Max(0, value - i);
// Add actual value with ans
ans = ans + current;
}
return ans;
}
// Driver code
public static int Main()
{
// Given array of values
int[] arr = { 6, 2, 4, 5 };
int N = arr.Length;
// Function call
Console.Write(maxSum(arr, N));
return 0;
}
}
// This code is contributed by Taranpreet
<script>
// JavaScript code for the above approach
// Find maximum possible sum
function maxSum(arr, N)
{
// Initialize ans with 0
let ans = 0;
// Sort array in descending order
arr.sort(function (a, b) { return b - a })
// Iterate array
for (let i = 0; i < N; i++) {
// Starting value
let value = arr[i];
// Actual value when being added
let current = Math.max(0, value - i);
// Add actual value with ans
ans = ans + current;
}
return ans;
}
// Driver code
// Given array of values
let arr = [6, 2, 4, 5];
let N = arr.length;
// Function call
document.write(maxSum(arr, N));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N log N)
Auxiliary Space: O(1)
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