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 |
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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 |
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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, 4a =[1, 5] b =[1, 1] if(check(n, k, a, b)): print("Yes") else: print("No") |
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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 |
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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 ?> |
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Output:
Yes
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