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Split the array into equal sum parts according to given conditions in C++
Here we will see one problem. Suppose one array arr is given. We have to check whether the array can be split into two parts, such that −
- Sub of both sub-arrays will be the same
- All elements, which are multiple of 5, will be in the same group
- All elements which are multiple of 3, but not multiple of 5, will be in the same group
- All other elements will be in other groups.
Suppose the array elements are {1, 4, 3}, then this can be split, because the sum of {1, 3} is the same as the sum of {4}, and the groups are also correct for the given condition.
Algorithm
isSplitArray(arr, n, start, left_sum, right_sum) −
Begin if start = n, then return true when left_sum = right_sum, otherwise false if arr[start] is divisible by 5, then add arr[start] with the left_sum else if arr[start] is divisible by 3, then add arr[start] with the right_sum else return isSplitArray(arr, n, start + 1, left_sum + arr[start], right_sum) OR isSplitArray(arr, n, start + 1, left_sum, right_sum + arr[start]) isSplitArray(arr, n, start + 1, left_sum, right_sum) End
Example
#include <iostream>
using namespace std;
bool isSplitArray(int* arr, int n, int start, int left_sum, int right_sum) {
if (start == n) //when it reaches at the end
return left_sum == right_sum;
if (arr[start] % 5 == 0) //when the element is divisible by 5, add to left sum
left_sum += arr[start];
else if (arr[start] % 3 == 0) //when the element is divisible by 3 but not 5, add to right sum
right_sum += arr[start];
else // otherwise it can be added to any of the sub-arrays
return isSplitArray(arr, n, start + 1, left_sum + arr[start], right_sum) || isSplitArray(arr, n, start + 1, left_sum, right_sum + arr[start]);
// For cases when element is multiple of 3 or 5.
return isSplitArray(arr, n, start + 1, left_sum, right_sum);
}
int main() {
int arr[] = {1, 4, 3};
int n = sizeof(arr)/sizeof(arr[0]);
if(isSplitArray(arr, n, 0, 0, 0)){
cout <<"Can be split";
} else {
cout <<"Can not be split";
}
}Output
Can be split
Updated on: 17-Oct-2019
647 Views
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