Given two sequences of integers ‘A’ and ‘B’, and an integer ‘k’. The task is to check if we can make both sequences equal by modifying any one element from the sequence A in the following way:
We can add any number from the range [-k, k] to any element of A. This operation must only be performed once. Print Yes if it is possible or No otherwise.
Examples:
Input: K = 2, A[] = {1, 2, 3}, B[] = {3, 2, 1}
Output: Yes
0 can be added to any element and both the sequences will be equal.Input: K = 4, A[] = {1, 5}, B[] = {1, 1}
Output: Yes
-4 can be added to 5 then the sequence A becomes {1, 1} which is equal to the sequence B.
Approach: Notice that to make both the sequence equal with just one move there has to be only one mismatching element in both the sequences and the absolute difference between them must be less than or equal to ‘k’.
- Sort both the arrays and look for the mismatching elements.
- If there are more than one mismatch elements then print ‘No’
- Else, find the absolute difference between the elements.
- If the difference <= k then print ‘Yes’ else print ‘No’.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include<bits/stdc++.h>using namespace std;// Function to check if both // sequences can be made equal static bool check(int n, int k, int *a, int *b) { // Sorting both the arrays sort(a,a+n); sort(b,b+n); // Flag to tell if there are // more than one mismatch bool fl = false; // To stores the index // of mismatched element int ind = -1; for (int i = 0; i < n; i++) { if (a[i] != b[i]) { // If there is more than one // mismatch then return False if (fl == true) { return false; } fl = true; ind = i; } } // If there is no mismatch or the // difference between the // mismatching elements is <= k // then return true if (ind == -1 | abs(a[ind] - b[ind]) <= k) { return true; } return false;}// Driver codeint main(){ int n = 2, k = 4; int a[] = {1, 5}; int b[] = {1, 1}; if (check(n, k, a, b)) { printf("Yes"); } else { printf("No"); } return 0;}// This code is contributed by mits |
Java
// Java implementation of the above approachimport java.util.Arrays;class GFG { // Function to check if both // sequences can be made equal static boolean check(int n, int k, int[] a, int[] b) { // Sorting both the arrays Arrays.sort(a); Arrays.sort(b); // Flag to tell if there are // more than one mismatch boolean fl = false; // To stores the index // of mismatched element int ind = -1; for (int i = 0; i < n; i++) { if (a[i] != b[i]) { // If there is more than one // mismatch then return False if (fl == true) { return false; } fl = true; ind = i; } } // If there is no mismatch or the // difference between the // mismatching elements is <= k // then return true if (ind == -1 | Math.abs(a[ind] - b[ind]) <= k) { return true; } return false; } // Driver code public static void main(String[] args) { int n = 2, k = 4; int[] a = {1, 5}; int b[] = {1, 1}; if (check(n, k, a, b)) { System.out.println("Yes"); } else { System.out.println("No"); } }} // This code is contributed by 29AjayKumar |
Python 3
# Python implementation of the above approach# Function to check if both # sequences can be made equaldef check(n, k, a, b): # Sorting both the arrays a.sort() b.sort() # Flag to tell if there are # more than one mismatch fl = False # To stores the index # of mismatched element ind = -1 for i in range(n): if(a[i] != b[i]): # If there is more than one # mismatch then return False if(fl == True): return False fl = True ind = i # If there is no mismatch or the # difference between the # mismatching elements is <= k # then return true if(ind == -1 or abs(a[ind]-b[ind]) <= k): return True return Falsen, k = 2, 4a =[1, 5]b =[1, 1]if(check(n, k, a, b)): print("Yes")else: print("No") |
C#
// C# implementation of the above approachusing System;class GFG { // Function to check if both // sequences can be made equal static bool check(int n, int k, int[] a, int[] b) { // Sorting both the arrays Array.Sort(a); Array.Sort(b); // Flag to tell if there are // more than one mismatch bool fl = false; // To stores the index // of mismatched element int ind = -1; for (int i = 0; i < n; i++) { if (a[i] != b[i]) { // If there is more than one // mismatch then return False if (fl == true) { return false; } fl = true; ind = i; } } // If there is no mismatch or the // difference between the // mismatching elements is <= k // then return true if (ind == -1 | Math.Abs(a[ind] - b[ind]) <= k) { return true; } return false; } // Driver code public static void Main() { int n = 2, k = 4; int[] a = {1, 5}; int[] b = {1, 1}; if (check(n, k, a, b)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript Implementation of above approach. // Function to check if both // sequences can be made equal function check(n, k, a, b) { // Sorting both the arrays a.sort(); b.sort(); // Flag to tell if there are // more than one mismatch let fl = false; // To stores the index // of mismatched element let ind = -1; for (let i = 0; i < n; i++) { if (a[i] != b[i]) { // If there is more than one // mismatch then return False if (fl == true) { return false; } fl = true; ind = i; } } // If there is no mismatch or the // difference between the // mismatching elements is <= k // then return true if (ind == -1 | Math.abs(a[ind] - b[ind]) <= k) { return true; } return false; } // Driver code let n = 2, k = 4; let a = [1, 5]; let b = [1, 1]; if (check(n, k, a, b)) { document.write("Yes"); } else { document.write("No"); } </script> |
PHP
<?php // PHP implementation of the // above approach// Function to check if both // sequences can be made equalfunction check($n, $k, &$a, &$b){ // Sorting both the arrays sort($a); sort($b); // Flag to tell if there are // more than one mismatch $fl = False; // To stores the index // of mismatched element $ind = -1; for ($i = 0; $i < $n; $i++) { if($a[$i] != $b[$i]) { // If there is more than one // mismatch then return False if($fl == True) return False; $fl = True; $ind = $i; } } // If there is no mismatch or the // difference between the // mismatching elements is <= k // then return true if($ind == -1 || abs($a[$ind] - $b[$ind]) <= $k) return True; return False; }// Driver Code$n = 2;$k = 4;$a = array(1, 5);$b = array(1, 1);if(check($n, $k, $a, $b)) echo "Yes";else echo "No"; // This code is contributed by ita_c?> |
Output
Yes
Complexity Analysis:
- Time Complexity: O(nlog(n))
- Auxiliary Space: O(1)
Approach: Hash Map
Steps:
- Initialize an empty hash map, freqMap.
- Iterate over each element in sequence A and update the frequencies of elements in freqMap.
- Iterate over each element in sequence B and decrement the frequencies of elements in freqMap.
- If all frequencies in freqMap are 0 or within the range [-k, k], print “Yes”.
- Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <iostream>#include <unordered_map>#include <vector>using namespace std;bool makeSequencesEqual(int k, const vector<int>& A, const vector<int>& B){ unordered_map<int, int> freqMap; // Update frequencies for sequence A for (int num : A) { freqMap[num]++; } // Decrement frequencies for sequence B for (int num : B) { freqMap[num]--; } for (const auto& pair : freqMap) { int num = pair.first; int freq = pair.second; // Check if frequencies are within the range [-k, k] if (freq != 0 && abs(freq) > k) { return false; } } return true;}// Driver Codeint main(){ int k = 2; vector<int> A = { 1, 2, 3 }; vector<int> B = { 3, 2, 1 }; // Check if sequences can be made equal if (makeSequencesEqual(k, A, B)) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0;} |
Java
// Java implementation of the above approachimport java.util.HashMap;import java.util.Map;import java.util.ArrayList;import java.util.List;public class GFG { public static boolean makeSequencesEqual(int k, List<Integer> A, List<Integer> B) { Map<Integer, Integer> freqMap = new HashMap<>(); // Update frequencies for sequence A for (int num : A) { freqMap.put(num, freqMap.getOrDefault(num, 0) + 1); } // Decrement frequencies for sequence B for (int num : B) { freqMap.put(num, freqMap.getOrDefault(num, 0) - 1); } for (Map.Entry<Integer, Integer> entry : freqMap.entrySet()) { int num = entry.getKey(); int freq = entry.getValue(); // Check if frequencies are within the range [-k, k] if (freq != 0 && Math.abs(freq) > k) { return false; } } return true; } // Driver Code public static void main(String[] args) { int k = 2; List<Integer> A = new ArrayList<>(); A.add(1); A.add(2); A.add(3); List<Integer> B = new ArrayList<>(); B.add(3); B.add(2); B.add(1); // Check if sequences can be made equal if (makeSequencesEqual(k, A, B)) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code is contributed by Vaibhav Nandan |
Python
def make_sequences_equal(k, A, B): freq_map = {} # Update frequencies for sequence A for num in A: freq_map[num] = freq_map.get(num, 0) + 1 # Decrement frequencies for sequence B for num in B: freq_map[num] = freq_map.get(num, 0) - 1 for num, freq in freq_map.items(): # Check if frequencies are within the range [-k, k] if freq != 0 and abs(freq) > k: return False return True# Driver Codeif __name__ == "__main__": k = 2 A = [1, 2, 3] B = [3, 2, 1] # Check if sequences can be made equal if make_sequences_equal(k, A, B): print("Yes") else: print("No") |
C#
using System;using System.Collections.Generic;class GFG { static bool MakeSequencesEqual(int k, List<int> A, List<int> B) { Dictionary<int, int> freqMap = new Dictionary<int, int>(); // Update frequencies for sequence A foreach(int num in A) { if (freqMap.ContainsKey(num)) freqMap[num]++; else freqMap[num] = 1; } // Decrement frequencies for sequence B foreach(int num in B) { if (freqMap.ContainsKey(num)) freqMap[num]--; else freqMap[num] = -1; } foreach(KeyValuePair<int, int> pair in freqMap) { int num = pair.Key; int freq = pair.Value; // Check if frequencies are within the range // [-k, k] if (freq != 0 && Math.Abs(freq) > k) { return false; } } return true; } // Driver Code static void Main() { int k = 2; List<int> A = new List<int>() { 1, 2, 3 }; List<int> B = new List<int>() { 3, 2, 1 }; // Check if sequences can be made equal if (MakeSequencesEqual(k, A, B)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }} |
Javascript
function makeSequencesEqual(k, A, B) { const freqMap = new Map(); // Update frequencies for sequence A for (let num of A) { freqMap.set(num, (freqMap.get(num) || 0) + 1); } // Decrement frequencies for sequence B for (let num of B) { freqMap.set(num, (freqMap.get(num) || 0) - 1); } for (let [num, freq] of freqMap) { // Check if frequencies are within the range [-k, k] if (freq !== 0 && Math.abs(freq) > k) { return false; } } return true;}// Driver Code const k = 2; const A = [1, 2, 3]; const B = [3, 2, 1]; // Check if sequences can be made equal if (makeSequencesEqual(k, A, B)) { console.log("Yes"); } else { console.log("No"); } |
Output
Yes
Time Complexity: O(n), where n is the total number of elements in sequences A and B.
Auxiliary Space: O(m), where m is the number of unique elements in sequences A and B.
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