C++ Greedy Approach N-array maximum sum














































C++ Greedy Approach N-array maximum sum



Maximum sum of increasing order elements from n arrays

Given n arrays of size m each. Find the maximum sum obtained by selecting a number from each array such that the elements selected from the i-th array are more than the element selected from (i-1)-th array. If maximum sum cannot be obtained then return 0.

Examples: 

Input : arr[][] = {{7, 1, 3, 4},
                   {4, 5, 2, 1},
                   {5, 9, 1, 8}}
Output : 18
Explanation :
We can select 4 from first array, 5 from 
second array and 9 from third array.

Input : arr[][] = {{9, 8, 7},
                   {6, 5, 4},
                   {3, 2, 1}}
Output : 0
The idea is to start picking from the last array. We pick the maximum element from the last array, then we move to the second last array. In the second last array, we find the largest element which is smaller than the maximum element picked from the last array. We repeat this process until we reach the first array.
To obtain maximum sum we can sort all arrays and start bottom to up traversing each array from right to left and choose a number such that it is greater than the previous element. If we are not able to select an element from the array then return 0.
Below is C++ implementation of above approach
// CPP program to find maximum sum
// by selecting a element from n arrays
#include <bits/stdc++.h>
#define M 4
using namespace std;

// To calculate maximum sum by
// selecting element from each array
int maximumSum(int a[][M], int n) {

// Sort each array
for (int i = 0; i < n; i++)
    sort(a[i], a[i] + M);

// Store maximum element
// of last array
int sum = a[n - 1][M - 1];
int prev = a[n - 1][M - 1];
int i, j;

// Selecting maximum element from
// previoulsy selected element
for (i = n - 2; i >= 0; i--) {
    for (j = M - 1; j >= 0; j--) {
    if (a[i][j] < prev) {
        prev = a[i][j];
        sum += prev;
        break;
    }
    }

    // j = -1 means no element is
    // found in a[i] so return 0
    if (j == -1)
    return 0;
}

return sum;
}

// Driver program to test maximumSum
int main() {
int arr[][M] = {{1, 7, 3, 4},
                {4, 2, 5, 1},
                {9, 5, 1, 8}};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximium sum obtained="<<maximumSum(arr, n);
return 0;
}

-----------------------------------------------------------
Output:
Maximium sum obtained=18


The Worst-Case Time Complexity: O(mn Log m)
We can optimize the above solution to work in O(mn). We can skip sorting to find the maximum elements.
Below is C++ implementation of optimized approach:
// CPP program to find maximum sum
// by selecting a element from n arrays
#include <bits/stdc++.h>
#define M 4
using namespace std;

// To calculate maximum sum by
// selecting element from each array
int maximumSum(int a[][M], int n) {

// Store maximum element of last array
int prev = *max_element(&a[n-1][0],
                &a[n-1][M-1] + 1);

// Selecting maximum element from
// previoulsy selected element
int sum = prev;
for (int i = n - 2; i >= 0; i--) {

    int max_smaller = INT_MIN;
    for (int j = M - 1; j >= 0; j--) {
    if (a[i][j] < prev &&
        a[i][j] > max_smaller)
        max_smaller = a[i][j];
    }

    // max_smaller equals to INT_MIN means
    // no element is found in a[i] so
    // return 0
    if (max_smaller == INT_MIN)
    return 0;

    prev = max_smaller;
    sum += max_smaller;
}

return sum;
}

// Driver program to test maximumSum
int main() {
int arr[][M] = {{1, 7, 3, 4},
                {4, 2, 5, 1},
                {9, 5, 1, 8}};
int n = sizeof(arr) / sizeof(arr[0]);
cout <<"Maximum sum obtained="<<maximumSum(arr, n);
return 0;
}

----------------------------------------------------------
Output:
Maximum sum obtained=18


Time Complexity: O(mn)


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