Check whether it is possible to make both arrays equal by modifying a single element

Last Updated : 05 May, 2025

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 code int 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 approach import 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

# Python implementation of the above approach  # Function to check if both  # sequences can be made equal def 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 False  n, k = 2, 4 a =[1, 5] b =[1, 1] if(check(n, k, a, b)):     print("Yes") else:     print("No")

// C# implementation of the above approach using 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 equal function 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 Code int 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 approach import 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 Code if __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")

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.

Find if it is possible to make all elements of an array equal by the given operations

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Check whether it is possible to make both arrays equal by modifying a single element

Approach: Hash Map

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