# Minimum number of pairs required to be selected to cover a given range

Last Updated : 15 Feb, 2023

Given an array arr[] consisting of N pairs and a positive integer K, the task is to select the minimum number of pairs such that it covers the entire range [0, K]. If it is not possible to cover the range [0, K], then print -1.

Examples :

Input: arr[] = {{0, 3}, {2, 5}, {5, 6}, {6, 8}, {6, 10}}, K = 10
Output: 4
Explanation: One of the optimal ways is to select the ranges {{0, 3}, {2, 5}, {5, 6}, {6, 10}} which covers the entire range [0, 10].

Input: arr[] = {{0, 4}, {1, 5}, {5, 6}, {8, 10}}, N = 10
Output : -1

Naive Approach: The simplest approach to solve the problem is to generate all possible subsets and check for each subset, if it covers the entire range or not.

Time Complexity: O(2N Ã— N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by sorting the array in increasing order on the basis of the left element and for pairs with equal left elements, sort the elements in decreasing order of their right element. Now, choose the pairs optimally.

Follow the steps below to solve the problem:

• Sort the vector of pairs arr[] in increasing order of left element and if left elements are equal, then sort the array in decreasing order of right element of pair.
• Check if arr[0][0] != 0 then print -1.
• Initialize a variable, say R as arr[0][1] which stores the right bound of the range and ans as 1 which stores number of ranges needed to cover the range [0,  K].
• Iterate in the range [0, N-1] using the variable i and perform the following steps:
• If R == K, terminate the loop.
• Initialize a variable, say maxr as R which stores the maximum right bound where left bound ? R.
• Iterate in the range [i+1, N-1] using the variable j and perform the following steps:
• If arr[j][0] ? R, then modify the value of maxr as max(maxr, arr[j][1]).
• Modify the value of i as j-1 and the value of R as maxr and increment the value of ans by 1.
• If R is not equal to K, then print -1.
• Otherwise, print the value of ans.

Below is the implementation of the above approach:

## C++14

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to sort the elements in` `// increasing order of left element and` `// sort pairs with equal left element in` `// decreasing order of right element` `static` `bool` `cmp(pair<``int``, ``int``> a, pair<``int``, ``int``> b)` `{` `    ``if` `(a.first == b.first) {` `        ``return` `a.second > b.second;` `    ``}` `    ``else` `{` `        ``return` `b.first > a.first;` `    ``}` `}`   `// Function to select minimum number of pairs` `// such that it covers the entire range [0, K]` `int` `MinimumPairs(vector > arr, ``int` `K)` `{` `    ``int` `N = arr.size();`   `    ``// Sort the array using comparator` `    ``sort(arr.begin(), arr.end(), cmp);`   `    ``// If the start element` `    ``// is not equal to 0` `    ``if` `(arr[0].first != 0) {` `        ``return` `-1;` `    ``}`   `    ``// Stores the minimum` `    ``// number of pairs required` `    ``int` `ans = 1;`   `    ``// Stores the right bound of the range` `    ``int` `R = arr[0].second;`   `    ``// Iterate in the range[0, N-1]` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``if` `(R == K) {` `            ``break``;` `        ``}`   `        ``// Stores the maximum right bound` `        ``// where left bound ? R` `        ``int` `maxr = R;`   `        ``int` `j;`   `        ``// Iterate in the range [i+1, N-1]` `        ``for` `(j = i + 1; j < N; j++) {`   `            ``// If the starting point of j-th` `            ``// element is less than equal to R` `            ``if` `(arr[j].first <= R) {`   `                ``maxr = max(maxr, arr[j].second);` `            ``}`   `            ``// Otherwise` `            ``else` `{` `                ``break``;` `            ``}` `        ``}`   `        ``i = j - 1;` `        ``R = maxr;` `        ``ans++;` `    ``}`   `    ``// If the right bound` `    ``// is not equal to K` `    ``if` `(R != K) {` `        ``return` `-1;` `    ``}`   `    ``// Otherwise` `    ``else` `{` `        ``return` `ans;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Input` `    ``int` `K = 4;` `    ``vector > arr{ { 0, 6 } };`   `    ``// Function Call` `    ``cout << MinimumPairs(arr, K);` `}`

## Java

 `// Java Program for above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `// User defined Pair class` `class` `Pair ` `{` `    ``int` `x;` `    ``int` `y;`   `    ``// Constructor` `    ``public` `Pair(``int` `x, ``int` `y)` `    ``{` `        ``this``.x = x;` `        ``this``.y = y;` `    ``}` `}` `// class to sort the elements in` `// increasing order of left element and` `// sort pairs with equal left element in` `// decreasing order of right element` `class` `cmp ``implements` `Comparator ` `{`   `    ``public` `int` `compare(Pair p1, Pair p2)` `    ``{` `        ``if` `(p1.x == p2.x) ` `        ``{` `            ``return` `p1.y - p2.y;` `        ``}` `        ``else` `        ``{` `            ``return` `p2.x - p1.x;` `        ``}` `    ``}` `}`   `class` `GFG ` `{` `    ``// Function to select minimum number of pairs` `    ``// such that it covers the entire range [0, K]` `    ``public` `static` `int` `MinimumPairs(Pair arr[], ``int` `K)` `    ``{` `        ``int` `N = arr.length;` `        ``// Sort the array using comparator` `        ``Arrays.sort(arr, ``new` `cmp());`   `        ``// If the start element` `        ``// is not equal to 0` `        ``if` `(arr[``0``].x != ``0``) ` `        ``{` `            ``return` `-``1``;` `        ``}`   `        ``// Stores the minimum` `        ``// number of pairs required` `        ``int` `ans = ``1``;`   `        ``// Stores the right bound of the range` `        ``int` `R = arr[``0``].y;`   `        ``// Iterate in the range[0, N-1]` `        ``for` `(``int` `i = ``0``; i < N; i++) ` `        ``{` `            ``if` `(R == K) ` `            ``{` `                ``break``;` `            ``}`   `            ``// Stores the maximum right bound` `            ``// where left bound is less than equal to R` `            ``int` `maxr = R;` `            ``int` `j;`   `            ``// Iterate in the range [i+1, N-1]` `            ``for` `(j = i + ``1``; j < N; j++) ` `            ``{`   `                ``// If the starting point of j-th` `                ``// element is less than equal to R` `                ``if` `(arr[j].x <= R) ` `                ``{`   `                    ``maxr = Math.max(maxr, arr[j].y);` `                ``}`   `                ``// Otherwise` `                ``else` `                ``{` `                    ``break``;` `                ``}` `            ``}`   `            ``i = j - ``1``;` `            ``R = maxr;` `            ``ans++;` `        ``}`   `        ``// If the right bound` `        ``// is not equal to K` `        ``if` `(R != K) ` `        ``{` `            ``return` `-``1``;` `        ``}`   `        ``// Otherwise` `        ``else` `        ``{` `            ``return` `ans;` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `K = ``4``;` `        ``int` `n = ``1``; ``// length of array` `        ``Pair arr[] = ``new` `Pair[n];` `        ``arr[``0``] = ``new` `Pair(``0``, ``6``);`   `        ``System.out.println(MinimumPairs(arr, K));` `    ``}` `}`   `// This code is contributed by rj13to.`

## Python3

 `# Python3 program for the above approach`   `# Function to select minimum number of pairs` `# such that it covers the entire range [0, K]` `def` `MinimumPairs(arr, K):` `    `  `    ``N ``=` `len``(arr)`   `    ``# Sort the array using comparator` `    ``arr.sort()`   `    ``# If the start element` `    ``# is not equal to 0` `    ``if` `(arr[``0``][``0``] !``=` `0``):` `        ``return` `-``1`   `    ``# Stores the minimum` `    ``# number of pairs required` `    ``ans ``=` `1`   `    ``# Stores the right bound of the range` `    ``R ``=` `arr[``0``][``1``]`   `    ``# Iterate in the range[0, N-1]` `    ``for` `i ``in` `range``(N):` `        ``if` `(R ``=``=` `K):` `            ``break`   `        ``# Stores the maximum right bound` `        ``# where left bound ? R` `        ``maxr ``=` `R`   `        ``j ``=` `0`   `        ``# Iterate in the range [i+1, N-1]` `        ``for` `j ``in` `range``(i ``+` `1``, N, ``1``):` `            `  `            ``# If the starting point of j-th` `            ``# element is less than equal to R` `            ``if` `(arr[j][``0``] <``=` `R):` `                ``maxr ``=` `max``(maxr, arr[j][``1``])`   `            ``# Otherwise` `            ``else``:` `                ``break`   `        ``i ``=` `j ``-` `1` `        ``R ``=` `maxr` `        ``ans ``+``=` `1`   `    ``# If the right bound` `    ``# is not equal to K` `    ``if` `(R !``=` `K):` `        ``return` `-``1`   `    ``# Otherwise` `    ``else``:` `        ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Given Input` `    ``K ``=` `4` `    ``arr ``=` `[ [ ``0``, ``6` `] ]`   `    ``# Function Call` `    ``print``(MinimumPairs(arr, K))`   `# This code is contributed by bgangwar59`

## Javascript

 ``

## C#

 `// C# program for the above approach`   `using` `System;` `using` `System.Linq;` `using` `System.Collections.Generic;`   `// User defined Pair class` `public` `class` `Pair` `{` `    ``public` `int` `x;` `    ``public` `int` `y;`     `    ``// Constructor` `    ``public` `Pair(``int` `x, ``int` `y)` `    ``{` `        ``this``.x = x;` `        ``this``.y = y;` `    ``}` `}`   `// class to sort the elements in` `// increasing order of left element and` `// sort pairs with equal left element in` `// decreasing order of right element` `public` `class` `cmp : IComparer` `{`   `    ``public` `int` `Compare(Pair p1, Pair p2)` `    ``{` `        ``if` `(p1.x == p2.x)` `        ``{` `            ``return` `p2.y - p1.y;` `        ``}` `        ``else` `        ``{` `            ``return` `p1.x - p2.x;` `        ``}` `    ``}` `}`     `class` `GFG` `{` `    ``// Function to select minimum number of pairs` `    ``// such that it covers the entire range [0, K]` `    ``public` `static` `int` `MinimumPairs(Pair[] arr, ``int` `K)` `    ``{` `        ``int` `N = arr.Length;` `        ``// Sort the array using comparator` `        ``Array.Sort(arr, ``new` `cmp());` `    `  `    `  `        ``// If the start element` `        ``// is not equal to 0` `        ``if` `(arr[0].x != 0)` `        ``{` `            ``return` `-1;` `        ``}` `    `  `        ``// Stores the minimum` `        ``// number of pairs required` `        ``int` `ans = 1;` `    `  `        ``// Stores the right bound of the range` `        ``int` `R = arr[0].y;` `    `  `        ``// Iterate in the range[0, N-1]` `        ``for` `(``int` `i = 0; i < N; i++)` `        ``{` `            ``if` `(R == K)` `            ``{` `                ``break``;` `            ``}` `    `  `            ``// Stores the maximum right bound` `            ``// where left bound is less than equal to R` `            ``int` `maxr = R;` `            ``int` `j;` `    `  `            ``// Iterate in the range [i+1, N-1]` `            ``for` `(j = i + 1; j < N; j++)` `            ``{` `    `  `                ``// If the starting point of j-th` `                ``// element is less than equal to R` `                ``if` `(arr[j].x <= R)` `                ``{` `    `  `                    ``maxr = Math.Max(maxr, arr[j].y);` `                ``}` `    `  `                ``// Otherwise` `                ``else` `                ``{` `                    ``break``;` `                ``}` `            ``}` `    `  `            ``i = j - 1;` `            ``R = maxr;` `            ``ans++;` `        ``}` `    `  `        ``// If the right bound` `        ``// is not equal to K` `        ``if` `(R != K)` `        ``{` `            ``return` `-1;` `        ``}` `    `  `        ``// Otherwise` `        ``else` `        ``{` `            ``return` `ans;` `        ``}` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int` `K = 4;` `        ``int` `n = 1; ``// length of array` `        ``Pair[] arr = ``new` `Pair[n];` `        ``arr[0] = ``new` `Pair(0, 6);` `    `  `        ``Console.WriteLine(MinimumPairs(arr, K));` `    ``}` `}`   `// This code is contributed by Harshad`

Output

`-1`

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

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