Maximum subarray sum in O(n) using prefix sum

08 Apr, 2021
Given an Array of Positive and Negative Integers, find out the Maximum Subarray Sum in that Array. 
Examples:  
Input1 : arr = {-2, -3, 4, -1, -2, 1, 5, -3}
Output1 : 7

Input2 : arr = {4, -8, 9, -4, 1, -8, -1, 6}
Output2 : 9
 
Kadane’s Algorithm solves this problem using Dynamic Programming approach in linear time. Here is another approach using Dynamic Programming and Prefix Sum to find out maximum subarray sum in Linear time.Prerequisite: Prefix Sum Array 
1. First calculate the prefix sum (prefix_sum) of the input array. 2. Sum of a subarray from index x to y can be presented as, 
    
3. Now maximum of these subarrays is, 
    
That is, we keep track of minimum prefix sum for x <= y and maximum subarray sum so far.  
Implementation: 1. Calculate the prefix sum of the input array. 2. Initialize : min_prefix_sum = 0, res = -infinite 3. Maintain a loop for i = 0 to n. (n is the size of the input array).      a) cand = prefix_sum[i] – mini     b) If cand is greater than res (maximum subarray sum so far), then update res by cand.     c) If prefix_sum[i] is less than min_prefix_sum (minimum prefix sum so far), then update min_prefix_sum by prefix_sum[i].4. Return res.Reference: Algorithms for the problem of k maximum sums and a VLSI algorithm for the k maximum subarrays problem 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find out maximum subarray// sum in linear time using prefix sum.#include <iostream>#include <limits>using namespace std; // Function to compute maximum subarray// sum in linear time.int maximumSumSubarray(int arr[], int n){    // Initialize minimum prefix sum to 0.    int min_prefix_sum = 0;     // Initialize maximum subarray sum so    // far to -infinity.    int res = numeric_limits<int>::min();     // Initialize and compute the prefix    // sum array.    int prefix_sum[n];    prefix_sum[0] = arr[0];    for (int i = 1; i < n; i++)        prefix_sum[i] = prefix_sum[i - 1] + arr[i];            // loop through the array, keep track    // of minimum prefix sum so far and    // maximum subarray sum.    for (int i = 0; i < n; i++) {        res = max(res, prefix_sum[i] -                       min_prefix_sum);        min_prefix_sum = min(min_prefix_sum,                             prefix_sum[i]);    }         return res;} // Driver Programint main(){    // Test case 1    int arr1[] = { -2, -3, 4, -1, -2, 1, 5, -3 };    int n1 = sizeof(arr1) / sizeof(arr1[0]);    cout << maximumSumSubarray(arr1, n1) << endl;     // Test case 2    int arr2[] = { 4, -8, 9, -4, 1, -8, -1, 6 };    int n2 = sizeof(arr2) / sizeof(arr2[0]);    cout << maximumSumSubarray(arr2, n2);     return 0;}
// Java program to find// out maximum subarray// sum in linear time// using prefix sum. class GFG{    // Function to compute maximum    // subarray sum in linear time.    static int maximumSumSubarray(int arr[], int n)    {        // Initialize minimum        // prefix sum to 0.        int min_prefix_sum = 0;             // Initialize maximum subarray        // sum so far to -infinity.        int res = Integer.MIN_VALUE;             // Initialize and compute        // the prefix sum array.        int prefix_sum[] = new int[n];        prefix_sum[0] = arr[0];        for (int i = 1; i < n; i++)            prefix_sum[i] = prefix_sum[i - 1]                            + arr[i];                 // loop through the array, keep        // track of minimum prefix sum so        // far and maximum subarray sum.        for (int i = 0; i < n; i++)        {            res = Math.max(res, prefix_sum[i] -                           min_prefix_sum);            min_prefix_sum = Math.min(min_prefix_sum,                                     prefix_sum[i]);        }                 return res;    }         // Driver Program    public static void main (String[] args)    {        // Test case 1        int arr1[] = { -2, -3, 4, -1, -2, 1, 5, -3 };        int n1 = arr1.length;        System.out.println(maximumSumSubarray(arr1, n1));             // Test case 2        int arr2[] = { 4, -8, 9, -4, 1, -8, -1, 6 };        int n2 = arr2.length;        System.out.println(maximumSumSubarray(arr2, n2));    }} // This code is contributed by Ansu Kumari.
# Python3 program to find out# maximum subarray sum in# linear time using prefix# sum.import math # Function to compute maximum# subarray sum in linear time.def maximumSumSubarray(arr, n):         # Initialize minimum prefix    # sum to 0.    min_prefix_sum = 0     # Initialize maximum subarray    # sum so far to -infinity.    res = -math.inf     # Initialize and compute the    # prefix sum array.    prefix_sum = []    prefix_sum.append(arr[0])         for i in range(1, n):        prefix_sum.append(prefix_sum[i - 1] + arr[i])         # loop through the array keep    # track of minimum prefix sum    # so far and maximum subarray    # sum.    for i in range(n):                 res = max(res, prefix_sum[i] - min_prefix_sum)        min_prefix_sum = min(min_prefix_sum, prefix_sum[i])         return res # Driver Program # Test case 1arr1 = [ -2, -3, 4, -1, -2, 1, 5, -3 ]n1 = len(arr1)print(maximumSumSubarray(arr1, n1)) # Test case 2arr2 = [ 4, -8, 9, -4, 1, -8, -1, 6 ]n2 = len(arr2)print(maximumSumSubarray(arr2, n2)) # This code is contributed by Ansu Kuamri.
// C# program to find// out maximum subarray// sum in linear time// using prefix sum.using System; class GFG{    // Function to compute maximum    // subarray sum in linear time.    static int maximumSumSubarray(int []arr, int n)    {        // Initialize minimum        // prefix sum to 0.        int min_prefix_sum = 0;             // Initialize maximum subarray        // sum so far to -infinity.        int res = int.MinValue;             // Initialize and compute        // the prefix sum array.        int []prefix_sum = new int[n];        prefix_sum[0] = arr[0];        for (int i = 1; i < n; i++)            prefix_sum[i] = prefix_sum[i - 1]                            + arr[i];             // loop through the array, keep        // track of minimum prefix sum so        // far and maximum subarray sum.        for (int i = 0; i < n; i++)        {            res = Math.Max(res, prefix_sum[i] -                        min_prefix_sum);            min_prefix_sum = Math.Min(min_prefix_sum,                                    prefix_sum[i]);        }                 return res;    }         // Driver Program    public static void Main ()    {        // Test case 1        int []arr1 = { -2, -3, 4, -1, -2, 1, 5, -3 };        int n1 = arr1.Length;        Console.WriteLine(maximumSumSubarray(arr1, n1));             // Test case 2        int []arr2 = { 4, -8, 9, -4, 1, -8, -1, 6 };        int n2 = arr2.Length;        Console.WriteLine(maximumSumSubarray(arr2, n2));    }} // This code is contributed by vt_m.
<?php// PHP program to find out// maximum subarray sum in// linear time using prefix sum. // Function to compute maximum// subarray sum in linear time.function maximumSumSubarray($arr, $n){    // Initialize minimum    // prefix sum to 0.    $min_prefix_sum = 0;     // Initialize maximum subarray    // sum so far to -infinity.    $res = PHP_INT_MIN;     // Initialize and compute    // the prefix sum array.    $prefix_sum = array();    $prefix_sum[0] = $arr[0];    for ($i = 1; $i < $n; $i++)        $prefix_sum[$i] = $prefix_sum[$i - 1] +                                      $arr[$i];     // loop through the array,    // keep track of minimum    // prefix sum so far and    // maximum subarray sum.    for ($i = 0; $i < $n; $i++)    {        $res = max($res, $prefix_sum[$i] -                         $min_prefix_sum);        $min_prefix_sum = min($min_prefix_sum,                              $prefix_sum[$i]);    }         return $res;} // Driver Code // Test case 1$arr1 = array(-2, -3, 4, -1,              -2, 1, 5, -3);$n1 = count($arr1);echo maximumSumSubarray($arr1, $n1), " \n" ; // Test case 2$arr2 = array(4, -8, 9, -4,              1, -8, -1, 6);               $n2 = count($arr2);echo maximumSumSubarray($arr2, $n2); // This code is contributed by anuj_67.?>
<script> // JavaScript program to find// out maximum subarray// sum in linear time// using prefix sum. // Function to compute maximum// subarray sum in linear time.    function maximumSumSubarray(arr, n)    {        // Initialize minimum        // prefix sum to 0.        let min_prefix_sum = 0;               // Initialize maximum subarray        // sum so far to -infinity.        let res = Number.MIN_VALUE;               // Initialize and compute        // the prefix sum array.        let prefix_sum = [];        prefix_sum[0] = arr[0];        for (let i = 1; i < n; i++)            prefix_sum[i] = prefix_sum[i - 1]                            + arr[i];                   // loop through the array, keep        // track of minimum prefix sum so        // far and maximum subarray sum.        for (let i = 0; i < n; i++)        {            res = Math.max(res, prefix_sum[i] -                           min_prefix_sum);            min_prefix_sum = Math.min(min_prefix_sum,                                     prefix_sum[i]);        }                   return res;    } // Driver code         // Test case 1        let arr1 = [ -2, -3, 4, -1, -2, 1, 5, -3 ];        let n1 = arr1.length;        document.write(maximumSumSubarray(arr1, n1) + "<br/>");               // Test case 2        let arr2 = [ 4, -8, 9, -4, 1, -8, -1, 6 ];        let n2 = arr2.length;        document.write(maximumSumSubarray(arr2, n2));            </script>
Output : 
7
9
Simpler and Efficient Solution Time Complexity: O(n). It takes linear time to compute the prefix sum and takes constant time in each iteration of the for loop. Hence overall complexity is O(n).Please note that the above problem can be solved in O(n) time and O(1) extra space using Kadane’s algorithm 
vt_m
AnshulAggarwal
susmitakundugoaldanga
prefix-sum
subarray
subarray-sum
Arrays
Dynamic Programming
prefix-sum
Arrays
Dynamic Programming
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Arrays in Java
Write a program to reverse an array or string
Maximum and minimum of an array using minimum number of comparisons
Top 50 Array Coding Problems for Interviews
Largest Sum Contiguous Subarray
Largest Sum Contiguous Subarray
Program for Fibonacci numbers
Longest Palindromic Substring | Set 1
Longest Increasing Subsequence | DP-3
Find if there is a path between two vertices in an undirected graph