The Maximum Subarray

Problem Statement :

We define subsequence as any subset of an array. We define a subarray as a contiguous subsequence in an array.

Given an array, find the maximum possible sum among:

1.all nonempty subarrays.
2.all nonempty subsequences.
Print the two values as space-separated integers on one line.

Note that empty subarrays/subsequences should not be considered.

arr = [-1, 2, 3, -4, 5,10]
The maximum subarray sum is comprised of elements at inidices [1-5]. Their sum is 2 + 3 + -4 +5 + 10 = 16. The maximum subsequence sum is comprised of elements at indices [1,2,4,5] and their sum is 2 + 3 + 5 + 10 = 20.

Function Description

Complete the maxSubarray function in the editor below.

maxSubarray has the following parameter(s):

int arr[n]: an array of integers

int[2]: the maximum subarray and subsequence sums
Input Format

The first line of input contains a single integer t, the number of test cases.

The first line of each test case contains na single integer n.
The second line contains  space-separated integers arr[i] where 0 <= i < n.


1 <= t <= 10
1 <= n <= 10^5
-10^4 <= arr[i] <= 10^4
The subarray and subsequences you consider should have at least one element.

Solution :


                            Solution in C :

In C++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int loop;
    cin >> loop;
    while (loop--) {
        int size;
        cin >> size;
        vector<int> data(size, 0);
        for (int i = 0; i < size; ++i) {
            cin >> data[i];

        vector<int> dp(size, 0);
        int big = 0;
        int sum = 0;
        int start = -1;
        for (int i = 0; i < size; i++) {
            int val = sum + data[i];

            if (val > 0) {
                if (sum == 0) {
                    start = i;
                sum = val;
            } else {
                sum = 0;

            if (sum > big) {
                big = sum;

        if (start == -1) {
            cout << data[0] << " ";
        } else {
            cout << big << " ";

        sum = 0;
        start = -1;
        for (int i = 0; i < size; ++i) {
            if (data[i] > 0) {
                start = i;
                sum += data[i];

        if (start == -1) {
            cout << data[0] << endl;
        } else {
            cout << sum << endl;
    return 0;

In Java :

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner myScan = new Scanner(;
        int T = myScan.nextInt();
        	int N = myScan.nextInt();
        	int[] arr = new int[N];
        	for(int i=0; i<N; i++){
        	int curr_sum=0;
        	int best_sum=0;
        	int non_cont_sum=0;
        	for(int j=0; j<N; j++){
        		curr_sum += arr[j];
        			non_cont_sum +=arr[j];
        	if(best_sum ==0 && non_cont_sum==0){
        		int max = arr[0];
        		for(int l=1; l<N; l++){
        	System.out.print(best_sum+" "+non_cont_sum);

In C :

#include <stdio.h>
#include <stdlib.h>
long long dp[100000];

int main(){
  long long a,T,N,ans1,ans2,max,i;
      if(i && dp[i-1]>0)
    printf("%lld %lld\n",ans1,ans2);
  return 0;

In Pyhton3 :

# your code goes here
test = int(input())
while test:
	test -= 1
	num_list = [int(x) for x in input().split()]
	max_sum = 0
	tmp_sum = 0
	max_non_contg = 0
	flg = False
	mx = max(num_list)
	if mx < 0:
		print(str(mx) + " " + str(mx))
	for i in range(len(num_list)):
		if tmp_sum < 0:
			tmp_sum = 0
		tmp_sum += num_list[i]
		if tmp_sum > max_sum:
			max_sum = tmp_sum
		if num_list[i] > 0:
			max_non_contg += num_list[i]
	print(str(max_sum) + " " + str(max_non_contg))

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