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Check whether it is possible to make both arrays equal by modifying a single element
• Difficulty Level : Basic
• Last Updated : 21 May, 2021

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#includeusing namespace std; // Function to check if both// sequences can be made equalstatic 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 False n, 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


Output:
Yes

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