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))
View More Similar Problems
Tree: Height of a Binary Tree
The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary
View Solution →Tree : Top View
Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.
View Solution →Tree: Level Order Traversal
Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F
View Solution →Binary Search Tree : Insertion
You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <
View Solution →Tree: Huffman Decoding
Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t
View Solution →Binary Search Tree : Lowest Common Ancestor
You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b
View Solution →