**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. Example 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 Returns 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. Constraints 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.io.*;
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(System.in);
int T = myScan.nextInt();
while(T--!=0){
int N = myScan.nextInt();
int[] arr = new int[N];
for(int i=0; i<N; i++){
arr[i]=myScan.nextInt();
}
int curr_sum=0;
int best_sum=0;
int non_cont_sum=0;
for(int j=0; j<N; j++){
curr_sum += arr[j];
if(curr_sum>0){
if(curr_sum>best_sum){
best_sum=curr_sum;
}
}else{
curr_sum=0;
}
if(arr[j]>0){
non_cont_sum +=arr[j];
}
}
if(best_sum ==0 && non_cont_sum==0){
int max = arr[0];
for(int l=1; l<N; l++){
if(arr[l]>max){
max=arr[l];
}
}
best_sum=max;
non_cont_sum=max;
}
System.out.print(best_sum+" "+non_cont_sum);
System.out.println();
}
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
long long dp[100000];
int main(){
long long a,T,N,ans1,ans2,max,i;
scanf("%lld",&T);
while(T--){
ans1=ans2=0;
max=-999999;
scanf("%lld",&N);
for(i=0;i<N;i++){
scanf("%lld",&a);
if(a>max)
max=a;
if(a>0)
ans2+=a;
dp[i]=a;
if(i && dp[i-1]>0)
dp[i]+=dp[i-1];
if(dp[i]>ans1)
ans1=dp[i];
}
if(ans1==0)
ans1=max;
if(ans2==0)
ans2=max;
printf("%lld %lld\n",ans1,ans2);
}
return 0;
}
In Pyhton3 :
# your code goes here
test = int(input())
while test:
test -= 1
input()
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))
continue
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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