Given N pairs of integers and an integer K, the task is to find the minimum number of reductions required such that the sum of the first elements of each pair is ≤ K.Each reduction involves reducing the first value of a pair to its second value. If it is not possible to make the sum ≤ K, print -1.
Examples:
Input: N = 5, K = 32
10 6
6 4
8 5
9 8
5 2
Output: 2
Explanation:
Total Sum = 10 + 6 + 8 + 9 + 5 = 38 > K
Reducing 10 – > 6 and 8 – > 5 reduces the sum to 31( 6 + 6 + 5 + 9 + 5) which is less than K.Input: N = 4, K = 25
10 5
20 9
12 10
4 2
Output: -1
Approach:
Follow the steps below to solve the problem:
- Calculate the sum of the first element of every pair. If the sum is already ≤ K, print 0.
- Sort the given pairs based on their difference.
- Count the number of differences of pairs that need to be added in non-increasing order to get the sum to be less than K.
- If the sum exceeds K after traversal of all pairs, print -1. Otherwise, print the count.
Below is the implementation of the above approach:
C++
// C++ Program to find the count of // minimum reductions required to // get the required sum K #include <bits/stdc++.h> using namespace std; // Function to return the count // of minimum reductions int countReductions( vector<pair<int, int> >& v, int K) { int sum = 0; for (auto i : v) { sum += i.first; } // If the sum is already // less than K if (sum <= K) { return 0; } // Sort in non-increasing // order of difference sort(v.begin(), v.end(), [&]( pair<int, int> a, pair<int, int> b) { return (a.first - a.second) > (b.first - b.second); }); int i = 0; while (sum > K && i < v.size()) { sum -= (v[i].first - v[i].second); i++; } if (sum <= K) return i; return -1; } // Driver Code int main() { int N = 4, K = 25; vector<pair<int, int> > v(N); v[0] = { 10, 5 }; v[1] = { 20, 9 }; v[2] = { 12, 10 }; v[3] = { 4, 2 }; // Function Call cout << countReductions(v, K) << endl; return 0; } |
-1
Time Complexity:O(NlogN)
Auxillary Space:O(1)
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