Given an array arr[][] containing pairs of integers in increasing order of GCDs, and an integer K, the task is to find a pair of integers whose GCD is at least K and is also the least among all possible GCDs exceeding K. If there exists no such pair, then print -1.
Examples:
Input: arr[][] = [(3, 6), (15, 30), (25, 75), (30, 120)], K = 16
Output: (25, 75)
Explanation:
The GCD of (25, 75) is 25 which is greater than 16 and least among all possible GCD.Input: arr[] = [(2, 5), (12, 36), (13, 26)], K = 14
Output: -1
Naive Approach: The simplest approach is to iterate over all the pairs of the given array and check for each pair, if its GCD exceeds K. From all such pairs, print the pair having the least GCD.
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
Efficient Approach: The idea is to observe that the array elements are sorted in increasing order of their GCD values of the pair so use Binary Search. Follow the steps below to solve the problem:
- Calculate the mid-value of the search space and check if GCD of arr[mid] > K.
- If it exceeds K, then update the answer and reduce the upper value of the search space to mid – 1.
- If the GCD of arr[mid] ? K, then increase the lower value of the search space to mid + 1.
- Continue the above process until the lower value is less than or equal to the upper value.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate// the GCD of two numbersint GCD(int a, int b){ if (b == 0) { return a; } return GCD(b, a % b);}// Function to print the pair// having a gcd value just greater// than the given integervoid GcdPair(vector<pair<int, int> > arr, int k){ // Initialize variables int lo = 0, hi = arr.size() - 1, mid; pair<int, int> ans; ans = make_pair(-1, 0); // Iterate until low less // than equal to high while (lo <= hi) { // Calculate mid mid = lo + (hi - lo) / 2; if (GCD(arr[mid].first, arr[mid].second) > k) { ans = arr[mid]; hi = mid - 1; } // Reducing the search space else lo = mid + 1; } // Print the answer if (ans.first == -1) cout << "-1"; else cout << "( " << ans.first << ", " << ans.second << " )"; return;}// Driver Codeint main(){ // Given array and K vector<pair<int, int> > arr = { { 3, 6 }, { 15, 30 }, { 25, 75 }, { 30, 120 } }; int K = 16; // Function Call GcdPair(arr, K); return 0;} |
Java
// Java program for // the above approachimport java.util.*;class GFG{// Function to calculate// the GCD of two numbersstatic int GCD(int a, int b){ if (b == 0) { return a; } return GCD(b, a % b);}// Function to print the pair// having a gcd value just // greater than the given integerstatic void GcdPair(int [][]arr, int k){ // Initialize variables int lo = 0, hi = arr.length - 1, mid; int []ans = {-1, 0}; // Iterate until low less // than equal to high while (lo <= hi) { // Calculate mid mid = lo + (hi - lo) / 2; if (GCD(arr[mid][0], arr[mid][1]) > k) { ans = arr[mid]; hi = mid - 1; } // Reducing the search space else lo = mid + 1; } // Print the answer if (ans[0] == -1) System.out.print("-1"); else System.out.print("( " + ans[0] + ", " + ans[1] + " )"); return;}// Driver Codepublic static void main(String[] args){ // Given array and K int [][]arr = {{3, 6}, {15, 30}, {25, 75}, {30, 120}}; int K = 16; // Function Call GcdPair(arr, K);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to calculate# the GCD of two numbersdef GCD(a, b): if (b == 0): return a return GCD(b, a % b)# Function to print the pair# having a gcd value just greater# than the given integerdef GcdPair(arr, k): # Initialize variables lo = 0 hi = len(arr) - 1 ans = [-1, 0] # Iterate until low less # than equal to high while (lo <= hi): # Calculate mid mid = lo + (hi - lo) // 2 if (GCD(arr[mid][0], arr[mid][1]) > k): ans = arr[mid] hi = mid - 1 # Reducing the search space else: lo = mid + 1 # Print the answer if (len(ans) == -1): print("-1") else: print("(", ans[0], ",", ans[1], ")")# Driver Codeif __name__ == '__main__': # Given array and K arr = [ [ 3, 6 ], [ 15, 30 ], [ 25, 75 ], [ 30, 120 ] ] K = 16 # Function call GcdPair(arr, K)# This code is contributed by mohit kumar 29 |
C#
// C# program for // the above approachusing System;class GFG{// Function to calculate// the GCD of two numbersstatic int GCD(int a, int b){ if (b == 0) { return a; } return GCD(b, a % b);}// Function to print the pair// having a gcd value just // greater than the given integerstatic void GcdPair(int [,]arr, int k){ // Initialize variables int lo = 0, hi = arr.Length - 1, mid; int []ans = {-1, 0}; // Iterate until low less // than equal to high while (lo <= hi) { // Calculate mid mid = lo + (hi - lo) / 2; if (GCD(arr[mid, 0], arr[mid, 1]) > k) { ans = GetRow(arr, mid); hi = mid - 1; } // Reducing the search space else lo = mid + 1; } // Print the answer if (ans[0] == -1) Console.Write("-1"); else Console.Write("( " + ans[0] + ", " + ans[1] + " )"); return;}public static int[] GetRow(int[,] matrix, int row){ var rowLength = matrix.GetLength(1); var rowVector = new int[rowLength]; for (var i = 0; i < rowLength; i++) rowVector[i] = matrix[row, i]; return rowVector;} // Driver Codepublic static void Main(String[] args){ // Given array and K int [,]arr = {{3, 6}, {15, 30}, {25, 75}, {30, 120}}; int K = 16; // Function Call GcdPair(arr, K);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program for // the above approach // Function to calculate // the GCD of two numbers function GCD(a , b) { if (b == 0) { return a; } return GCD(b, a % b); } // Function to print the pair // having a gcd value just // greater than the given integer function GcdPair(arr , k) { // Initialize variables var lo = 0, hi = arr.length - 1, mid; var ans = [ -1, 0 ]; // Iterate until low less // than equal to high while (lo <= hi) { // Calculate mid mid = lo + parseInt((hi - lo) / 2); if (GCD(arr[mid][0], arr[mid][1]) > k) { ans = arr[mid]; hi = mid - 1; } // Reducing the search space else lo = mid + 1; } // Print the answer if (ans[0] == -1) document.write("-1"); else document.write("( " + ans[0] + ", " + ans[1] + " )"); return; } // Driver Code // Given array and K var arr = [ [ 3, 6 ], [ 15, 30 ], [ 25, 75 ], [ 30, 120 ] ]; var K = 16; // Function Call GcdPair(arr, K);// This code is contributed by todaysgaurav</script> |
Output
( 25, 75 )
Time Complexity: O(log(N)2)
Auxiliary Space: O(1)
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