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 2Output:2Explanation:

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 2Output:-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;` `}` |

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## Java

`// Java program to find the count of ` `// minimum reductions required to ` `// get the required sum K ` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to return the count` `// of minimum reductions` `static` `int` `countReductions(ArrayList<` `int` `[]> v,` ` ` `int` `K)` `{` ` ` `int` `sum = ` `0` `;` ` ` `for` `(` `int` `[] i : v)` ` ` `{` ` ` `sum += i[` `0` `];` ` ` `}` ` ` `// If the sum is already` ` ` `// less than K` ` ` `if` `(sum <= K)` ` ` `{` ` ` `return` `0` `;` ` ` `}` ` ` `// Sort in non-increasing` ` ` `// order of difference` ` ` `Collections.sort(v, (a, b) -> Math.abs(b[` `0` `] - b[` `1` `]) -` ` ` `Math.abs(a[` `0` `] - a[` `1` `]));` ` ` `int` `i = ` `0` `;` ` ` `while` `(sum > K && i < v.size()) ` ` ` `{` ` ` `sum -= (v.get(i)[` `0` `] - v.get(i)[` `1` `]);` ` ` `i++;` ` ` `}` ` ` `if` `(sum <= K)` ` ` `return` `i;` ` ` `return` `-` `1` `;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `4` `, K = ` `25` `;` ` ` `ArrayList<` `int` `[]> v = ` `new` `ArrayList<>();` ` ` `v.add(` `new` `int` `[] { ` `10` `, ` `5` `});` ` ` `v.add(` `new` `int` `[] { ` `20` `, ` `9` `});` ` ` `v.add(` `new` `int` `[] { ` `12` `, ` `10` `});` ` ` `v.add(` `new` `int` `[] { ` `4` `, ` `2` `});` ` ` `// Function Call` ` ` `System.out.println(countReductions(v, K));` `}` `}` `// This code is contributed by offbeat` |

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## Python3

`# Python3 program to find the count of` `# minimum reductions required to` `# get the required sum K` `from` `typing ` `import` `Any` `, ` `List` `# Function to return the count` `# of minimum reductions` `def` `countReductions(v: ` `List` `[` `Any` `], K: ` `int` `) ` `-` `> ` `int` `:` ` ` `sum` `=` `0` ` ` ` ` `for` `i ` `in` `v:` ` ` `sum` `+` `=` `i[` `0` `]` ` ` `# If the sum is already` ` ` `# less than K` ` ` `if` `(` `sum` `<` `=` `K):` ` ` `return` `0` ` ` `# Sort in non-increasing` ` ` `# order of difference` ` ` `v.sort(key ` `=` `lambda` `a : a[` `0` `] ` `-` `a[` `1` `])` ` ` `i ` `=` `0` ` ` ` ` `while` `(` `sum` `> K ` `and` `i < ` `len` `(v)):` ` ` `sum` `-` `=` `(v[i][` `0` `] ` `-` `v[i][` `1` `])` ` ` `i ` `+` `=` `1` ` ` `if` `(` `sum` `<` `=` `K):` ` ` `return` `i` ` ` `return` `-` `1` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `4` ` ` `K ` `=` `25` ` ` `v ` `=` `[[` `0` `, ` `0` `] ` `for` `_ ` `in` `range` `(N)]` ` ` `v[` `0` `] ` `=` `[` `10` `, ` `5` `]` ` ` `v[` `1` `] ` `=` `[` `20` `, ` `9` `]` ` ` `v[` `2` `] ` `=` `[` `12` `, ` `10` `]` ` ` `v[` `3` `] ` `=` `[` `4` `, ` `2` `]` ` ` `# Function Call` ` ` `print` `(countReductions(v, K))` `# This code is contributed by sanjeev2552` |

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**Output:**

-1

**Time Complexity:**O(NlogN)**Auxiliary Space:**O(1)