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# Find minimum subarray sum for each index i in subarray [i, N-1]

• Difficulty Level : Hard
• Last Updated : 30 Jul, 2021

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 ``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 sart 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

 `// 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 sart 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

 `# 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 sart 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#

 `// 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 sart 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`

## Javascript

 ``
Output
`-5 -5 -2 4 4 `

Time Complexity: O(N^2)
Auxiliary Space: O(1)

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