Find minimum subarray sum for each index i in subarray [i, N-1]
Last Updated :
23 Jul, 2025
Given an array arr[] of size N, the task is to find the minimum subarray sum in the subarrays [i, N-1] for all i in [0, N-1].
Examples:
Input: arr[ ] = {3, -1, -2}
Output: -3 -3 -2
Explanation:
For (i = 1) i.e. {3, -1, -2}, the minimum subarray sum is -3 for {-1, -2}.
For (i = 2) i.e. {-1, -2}, the minimum subarray sum is -3 for {-1, -2}.
For (i = 3) i.e. {-2}, the minimum subarray sum is -2 for {-2}.
Input: arr[ ] = {5, -3, -2, 9, 4}
Output: -5 -5 -2 4 4
Approach: The problem can be solved by using the standard Kadane's algorithm for maximum subarray sum. Follow the steps below to solve this problem:
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Kadane's Algorithm to find max
// sum subarray
int kadane(int arr[], int start, int end)
{
int currMax = arr[start];
int maxSoFar = arr[start];
// Iterating from start to end
for (int i = start + 1; i < end + 1; i++) {
currMax = max(arr[i], arr[i] + currMax);
maxSoFar = max(maxSoFar, currMax);
}
// Returning maximum sum
return maxSoFar;
}
// Function to find the minimum subarray
// sum for each suffix
void minSubarraySum(int arr[], int n)
{
// Inverting all the elements of
// array arr[].
for (int i = 0; i < n; i++) {
arr[i] = -arr[i];
}
// Finding the result for each
// subarray
for (int i = 0; i < n; i++) {
// Finding the max subarray sum
int result = kadane(arr, i, n - 1);
// Inverting the result
result = -result;
// Print the result
cout << result << " ";
}
}
// Driver code
int main()
{
// Given Input
int n = 5;
int arr[] = { 5, -3, -2, 9, 4 };
// Function Call
minSubarraySum(arr, n);
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG{
// Kadane's Algorithm to find max
// sum subarray
static int kadane(int arr[], int start, int end)
{
int currMax = arr[start];
int maxSoFar = arr[start];
// Iterating from start to end
for(int i = start + 1; i < end + 1; i++)
{
currMax = Math.max(arr[i], arr[i] + currMax);
maxSoFar = Math.max(maxSoFar, currMax);
}
// Returning maximum sum
return maxSoFar;
}
// Function to find the minimum subarray
// sum for each suffix
static void minSubarraySum(int arr[], int n)
{
// Inverting all the elements of
// array arr[].
for(int i = 0; i < n; i++)
{
arr[i] = -arr[i];
}
// Finding the result for each
// subarray
for(int i = 0; i < n; i++)
{
// Finding the max subarray sum
int result = kadane(arr, i, n - 1);
// Inverting the result
result = -result;
// Print the result
System.out.print(result + " ");
}
}
// Driver code
public static void main(String[] args)
{
// Given Input
int n = 5;
int arr[] = { 5, -3, -2, 9, 4 };
// Function Call
minSubarraySum(arr, n);
}
}
// This code is contributed by Potta Lokesh
# Python3 program for the above approach
# Kadane's Algorithm to find max
# sum subarray
def kadane(arr, start, end):
currMax = arr[start]
maxSoFar = arr[start]
# Iterating from start to end
for i in range(start + 1,end + 1, 1):
currMax = max(arr[i], arr[i] + currMax)
maxSoFar = max(maxSoFar, currMax)
# Returning maximum sum
return maxSoFar
# Function to find the minimum subarray
# sum for each suffix
def minSubarraySum(arr, n):
# Inverting all the elements of
# array arr[].
for i in range(n):
arr[i] = -arr[i]
# Finding the result for each
# subarray
for i in range(n):
# Finding the max subarray sum
result = kadane(arr, i, n - 1)
# Inverting the result
result = -result
# Print the result
print(result, end = " ")
# Driver code
if __name__ == '__main__':
# Given Input
n = 5
arr = [ 5, -3, -2, 9, 4 ]
# Function Call
minSubarraySum(arr, n)
# This code is contributed by SURENDRA_GANGWAR
// C# program for the above approach
using System;
class GFG{
// Kadane's Algorithm to find max
// sum subarray
static int kadane(int []arr, int start, int end)
{
int currMax = arr[start];
int maxSoFar = arr[start];
// Iterating from start to end
for(int i = start + 1; i < end + 1; i++)
{
currMax = Math.Max(arr[i], arr[i] + currMax);
maxSoFar = Math.Max(maxSoFar, currMax);
}
// Returning maximum sum
return maxSoFar;
}
// Function to find the minimum subarray
// sum for each suffix
static void minSubarraySum(int []arr, int n)
{
// Inverting all the elements of
// array arr[].
for(int i = 0; i < n; i++)
{
arr[i] = -arr[i];
}
// Finding the result for each
// subarray
for(int i = 0; i < n; i++)
{
// Finding the max subarray sum
int result = kadane(arr, i, n - 1);
// Inverting the result
result = -result;
// Print the result
Console.Write(result + " ");
}
}
// Driver code
public static void Main(String[] args)
{
// Given Input
int n = 5;
int []arr = { 5, -3, -2, 9, 4 };
// Function Call
minSubarraySum(arr, n);
}
}
// This code is contributed by shivanisinghss2110
<script>
// Javascript program for the above approach
// Kadane's Algorithm to find max
// sum subarray
function kadane(arr, start, end)
{
let currMax = arr[start];
let maxSoFar = arr[start];
// Iterating from start to end
for(let i = start + 1; i < end + 1; i++)
{
currMax = Math.max(arr[i], arr[i] + currMax);
maxSoFar = Math.max(maxSoFar, currMax);
}
// Returning maximum sum
return maxSoFar;
}
// Function to find the minimum subarray
// sum for each suffix
function minSubarraySum(arr, n)
{
// Inverting all the elements of
// array arr[].
for(let i = 0; i < n; i++)
{
arr[i] = -arr[i];
}
// Finding the result for each
// subarray
for(let i = 0; i < n; i++)
{
// Finding the max subarray sum
let result = kadane(arr, i, n - 1);
// Inverting the result
result = -result;
// Print the result
document.write(result + " ");
}
}
// Driver code
// Given Input
let n = 5;
let arr = [ 5, -3, -2, 9, 4 ];
// Function Call
minSubarraySum(arr, n);
// This code is contributed by gfgking
</script>
Time Complexity: O(N^2)
Auxiliary Space: O(1)
Efficient Approach: The repetitive process of finding the maximum subarray sum using Kadane's can be optimized further by traversing the array from the back and storing the maximum negative sum till that point from the end.
Below is the code implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void minSubarraySum(int arr[], int n)
{
vector<int> res(n);
for (int i = 0; i < n; i++)
arr[i] *= -1;
int sum = 0;
int maxSum = INT_MIN;
for (int i = n - 1; i >= 0; i--) {
sum += arr[i];
maxSum = max(sum, maxSum);
// Store result for [i,n-1]
res[i] = -maxSum;
if (sum < 0)
sum = 0;
}
for (int i : res)
cout << i << " ";
}
int main()
{
int n = 5;
int arr[] = { 5, -3, -2, 9, 4 };
// Function Call
minSubarraySum(arr, n);
return 0;
}
// Java code to implement the approach
import java.io.*;
class GFG {
public static void minSubarraySum(int arr[], int n)
{
int[] res = new int[n];
for (int i = 0; i < n; i++)
arr[i] *= -1;
int sum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = n - 1; i >= 0; i--) {
sum += arr[i];
maxSum = Math.max(sum, maxSum);
// Store result for [i,n-1]
res[i] = -maxSum;
if (sum < 0)
sum = 0;
}
for (int i : res)
System.out.print(i + " ");
}
// Driver Code
public static void main(String[] args)
{
int n = 5;
int arr[] = { 5, -3, -2, 9, 4 };
// Function Call
minSubarraySum(arr, n);
}
}
// This code is contributed by aarohirai2616.
def minSubarraySum(arr, n):
res = [0]*n
for i in range(0, n):
arr[i] *= -1
suma = 0
maxSum = -9223372036854775808
i = n - 1
while(i >= 0):
suma += arr[i]
maxSum = max(suma, maxSum)
# Store result for [i,n-1]
res[i] = -maxSum
if (suma < 0):
suma = 0
i = i-1
for i in range(0, n):
print(res[i], end=" ")
# Driver code
n = 5
arr = [5, -3, -2, 9, 4]
# Function Call
minSubarraySum(arr, n)
# This code is contributed by aarohirai2616.
// C# code to implement the approach
using System;
public class GFG {
public static void minSubarraySum(int[] arr, int n)
{
int[] res = new int[n];
for (int i = 0; i < n; i++)
arr[i] *= -1;
int sum = 0;
int maxSum = Int32.MinValue;
for (int i = n - 1; i >= 0; i--) {
sum += arr[i];
maxSum = Math.Max(sum, maxSum);
// Store result for [i,n-1]
res[i] = -maxSum;
if (sum < 0)
sum = 0;
}
for (int i = 0; i < n; i++) {
Console.Write(res[i] + " ");
}
}
static public void Main()
{
// Code
int n = 5;
int[] arr = { 5, -3, -2, 9, 4 };
// Function Call
minSubarraySum(arr, n);
}
}
// This code is contributed by lokeshmvs21.
function minSubarraySum(arr, n)
{
let res = [];
for(let i=0;i<n;i++)
{
res.push(0);
}
for (let i = 0; i < n; i++)
arr[i] *= -1;
let sum = 0;
let maxSum = -2147483647 - 1;
for (let i = n - 1; i >= 0; i--) {
sum += arr[i];
maxSum = Math.max(sum, maxSum);
// Store result for [i,n-1]
res[i] = -1*maxSum;
if (sum < 0)
sum = 0;
}
console.log(res);
}
let n = 5;
let arr = [ 5, -3, -2, 9, 4 ];
// Function Call
minSubarraySum(arr, n);
// This code is contributed by akashish__
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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