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

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` `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 3

 `# 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#

 `// 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

 ``

## PHP

 ``

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 ` `#include ` `#include `   `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 A, List B) {` `        ``Map 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 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 A = ``new` `ArrayList<>();` `        ``A.add(``1``);` `        ``A.add(``2``);` `        ``A.add(``3``);` `        ``List 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"``)`

## 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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