# Maximum possible sum after M operations on N cards

Given an array arr[] of size N which represents the initial number on each card and given a two dimensional array B[][] of size M where M represents the number of operations that need to be performed. At each operation, choose at most B[j][0] cards (possibly zero) and replace the integer written on each chosen card with B[j][1]. The task is to find the maximum possible sum after M operations.

Examples:

Input: arr[] = {5, 1, 4}, B[][] = {{2, 3}, {1, 5}}
Output: 14
Replacing 1 with 5 and the sum becomes
5 + 5 + 4 = 14 which is the maximum possible.

Input: arr[] = {100, 100}, B[][] = {{2, 99}}
Output: 200

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: A greedy approach is applicable here. Sort the array arr[] in increasing order and sort the array B[][] in decreasing order of the number to be replaced. Then try to replace last non-replaced card of arr[] with one of the non-used cards of B[][]. Finally, print the maximized sum.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum ` `// possible sum after M operations ` `int` `max_sum(``int` `a[], ``int` `n, ``int` `b[][2], ``int` `m) ` `{ ` ` `  `    ``// Sort the array a in ` `    ``// increasing order ` `    ``sort(a, a + n); ` ` `  `    ``// Place all replacable cards in B ` `    ``vector > B; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``B.push_back({ b[i][1], b[i][0] }); ` ` `  `    ``// Sort vector B in decreasing order ` `    ``sort(B.rbegin(), B.rend()); ` ` `  `    ``// To store last unused card of a ` `    ``int` `left = 0; ` ` `  `    ``// Try to apply all m operations ` `    ``for` `(``int` `i = 0; i < m; i++) { ` `        ``int` `x = B[i].first, y = B[i].second; ` ` `  `        ``// Try for all applicable cards ` `        ``for` `(``int` `j = 0; j < y; j++) { ` ` `  `            ``// If current number on card is ` `            ``// less than applicable card ` `            ``if` `(a[left] < x) { ` `                ``a[left] = x; ` `                ``left++; ` ` `  `                ``if` `(left == n) ` `                    ``break``; ` `            ``} ` `            ``else` `                ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// To store the maximum ` `    ``// possible sum ` `    ``int` `ans = 0; ` ` `  `    ``// Calculate the maximum ` `    ``// possible sum ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans += a[i]; ` ` `  `    ``// Return the required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 5, 1, 4 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` `    ``int` `b[][2] = { { 2, 3 }, { 1, 5 } }; ` `    ``int` `m = ``sizeof``(b) / ``sizeof``(b[0]); ` ` `  `    ``cout << max_sum(a, n, b, m); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the maximum  ` `# possible sum after M operations  ` `def` `max_sum(a, n, b, m) : ` ` `  `    ``# Sort the array a in  ` `    ``# increasing order  ` `    ``a.sort();  ` ` `  `    ``# Place all replacable cards in B  ` `    ``B ``=` `[];  ` `    ``for` `i ``in` `range``(m) : ` `        ``B.append([b[i][``1``], b[i][``0``]]);  ` ` `  `    ``# Sort vector B in decreasing order  ` `    ``B.sort(reverse ``=` `True``)  ` ` `  `    ``# To store last unused card of a  ` `    ``left ``=` `0``;  ` ` `  `    ``# Try to apply all m operations  ` `    ``for` `i ``in` `range``(m) : ` `        ``x ``=` `B[i][``0``]; ` `        ``y ``=` `B[i][``1``];  ` ` `  `        ``# Try for all applicable cards  ` `        ``for` `j ``in` `range``(y) :  ` ` `  `            ``# If current number on card is  ` `            ``# less than applicable card  ` `            ``if` `(a[left] < x) : ` `                ``a[left] ``=` `x;  ` `                ``left ``+``=` `1``;  ` ` `  `                ``if` `(left ``=``=` `n) : ` `                    ``break``;  ` `            ``else` `: ` `                ``break``;  ` ` `  `    ``# To store the maximum  ` `    ``# possible sum  ` `    ``ans ``=` `0``;  ` ` `  `    ``# Calculate the maximum  ` `    ``# possible sum  ` `    ``for` `i ``in` `range``(n) : ` `        ``ans ``+``=` `a[i];  ` ` `  `    ``# Return the required answer  ` `    ``return` `ans;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``a ``=` `[``5``, ``1``, ``4``];  ` `    ``n ``=` `len``(a);  ` `    ``b ``=` `[[``2``, ``3``], [``1``, ``5``]];  ` `    ``m ``=` `len``(b);  ` ` `  `    ``print``(max_sum(a, n, b, m));  ` ` `  `# This code is contributed by AnkitRai01 `

Output:

```14
```

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