Kadane's Algorithm
PROBLEM :
Given an array containing both negative and positive integers. Find the contiguous sub-array with maximum sum.
Input:
The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Output:
Print the maximum sum of the contiguous sub-array in a separate line for each test case.
Constraints:
1 = T = 40
1 = N = 100
-100 = A[i] <= 100
Example:
Input
2
3
1 2 3
4
-1 -2 -3 -4
Output
6
-1
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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
int maxof(int a,int b){
return a>b?a:b ;
}
int main()
{
int t ;
cin>>t ;
while(t--){
int no ;
cin>>no ;
int arr[no] ;
for(int i=0;i<no;i++)
cin>>arr[i] ;
int sum,max ;
sum=arr[0] ;
max=arr[0] ;
for(int i=1;i<no;i++){
sum=maxof(arr[i],sum+arr[i]) ;
max=maxof(sum,max) ;
}
cout<<max<<endl ;
}
return 0;
}
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Given an array containing both negative and positive integers. Find the contiguous sub-array with maximum sum.
Input:
The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Output:
Print the maximum sum of the contiguous sub-array in a separate line for each test case.
Constraints:
1 = T = 40
1 = N = 100
-100 = A[i] <= 100
Example:
Input
2
3
1 2 3
4
-1 -2 -3 -4
Output
6
-1
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------
#include<iostream>
using namespace std;
int maxof(int a,int b){
return a>b?a:b ;
}
int main()
{
int t ;
cin>>t ;
while(t--){
int no ;
cin>>no ;
int arr[no] ;
for(int i=0;i<no;i++)
cin>>arr[i] ;
int sum,max ;
sum=arr[0] ;
max=arr[0] ;
for(int i=1;i<no;i++){
sum=maxof(arr[i],sum+arr[i]) ;
max=maxof(sum,max) ;
}
cout<<max<<endl ;
}
return 0;
}
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