Given two permutation arrays A[] and B[] of the first N Natural Numbers, the task is to find the minimum number of operations required to remove all array elements A[] such that in each operation remove the subsequence of array elements A[] whose order is the same as in the array B[].
Example:
Input: A[] = { 4, 2, 1, 3 }, B[] = { 1, 3, 2, 4 }
Output: 3
Explanation:
The given example can be solved by following the given steps:
- During the 1st operation, integers at index 2 and 3 in the array A[] can be deleted. Hence, the array A[] = {4, 2}.
- During the 2nd operation, integer at index 1 in the array A[] can be deleted. Hence, the array A[] = {4}.
- During the 3rd operation, integer at index 0 in the array A[] can be deleted. Hence, the array A[] = {}.
The order in which the elements are deleted is {1, 3, 2, 4} which is equal to B. Hence a minimum of 3 operations is required.
Input: A[] = {2, 4, 6, 1, 5, 3}, B[] = {6, 5, 4, 2, 3, 1}
Output: 4
Approach: The given problem can be solved using the steps discussed below:
- Create two variables i and j, where i keeps track of the index of the current element of B that is to be deleted next and j keeps track of the current element of A. Initially, both i=0 and j=0.
- Traverse the permutation array A[] using j for all values of j in range [0, N-1]. If A[j] = B[i], increment the value of i by 1 and continue traversing the array A[].
- After the array A[] has been traversed completely, increment the value of cnt variable which maintains the count of required operations.
- Repeat steps 2 and 3 till i<N.
- After completing the above steps, the value stored in cnt is the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum number of// operations to delete all elements of// permutation A in order described by Bint minOperations(int A[], int B[], int N){ // Stores the count of operations int cnt = 0; // Stores the index of current integer // in B to be deleted int i = 0; // Loop to iterate over all values of B while (i < N) { // Stores the current index in A int j = 0; // Iterate over all values A while (j < N && i < N) { // If current integer of B and A // equal, increment the index of // the current integer of B if (B[i] == A[j]) { i++; } j++; } // As the permutation A has been // traversed completely, increment // the count of operations by 1 cnt++; } // Return Answer return cnt;}// Driver Codeint main(){ int A[] = { 2, 4, 6, 1, 5, 3 }; int B[] = { 6, 5, 4, 2, 3, 1 }; int N = sizeof(A) / sizeof(A[0]); cout << minOperations(A, B, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the minimum number of// operations to delete all elements of// permutation A in order described by Bstatic int minOperations(int A[], int B[], int N){ // Stores the count of operations int cnt = 0; // Stores the index of current integer // in B to be deleted int i = 0; // Loop to iterate over all values of B while (i < N) { // Stores the current index in A int j = 0; // Iterate over all values A while (j < N && i < N) { // If current integer of B and A // equal, increment the index of // the current integer of B if (B[i] == A[j]) { i++; } j++; } // As the permutation A has been // traversed completely, increment // the count of operations by 1 cnt++; } // Return Answer return cnt;}// Driver Codepublic static void main(String[] args){ int A[] = { 2, 4, 6, 1, 5, 3 }; int B[] = { 6, 5, 4, 2, 3, 1 }; int N = A.length; System.out.print(minOperations(A, B, N));}}// This code is contributed by 29AjayKumar |
Python3
# Python 3 program for the above approach# Function to find the minimum number of# operations to delete all elements of# permutation A in order described by Bdef minOperations(A, B, N): # Stores the count of operations cnt = 0 # Stores the index of current integer # in B to be deleted i = 0 # Loop to iterate over all values of B while(i < N): # Stores the current index in A j = 0 # Iterate over all values A while (j < N and i < N): # If current integer of B and A # equal, increment the index of # the current integer of B if (B[i] == A[j]): i += 1 j += 1 # As the permutation A has been # traversed completely, increment # the count of operations by 1 cnt += 1 # Return Answer return cnt# Driver Codeif __name__ == '__main__': A = [2, 4, 6, 1, 5, 3] B = [6, 5, 4, 2, 3, 1] N = len(A) print(minOperations(A, B, N)) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;class GFG { // Function to find the minimum number of // operations to delete all elements of // permutation A in order described by B static int minOperations(int[] A, int[] B, int N) { // Stores the count of operations int cnt = 0; // Stores the index of current integer // in B to be deleted int i = 0; // Loop to iterate over all values of B while (i < N) { // Stores the current index in A int j = 0; // Iterate over all values A while (j < N && i < N) { // If current integer of B and A // equal, increment the index of // the current integer of B if (B[i] == A[j]) { i++; } j++; } // As the permutation A has been // traversed completely, increment // the count of operations by 1 cnt++; } // Return Answer return cnt; } // Driver Code public static void Main(string[] args) { int[] A = { 2, 4, 6, 1, 5, 3 }; int[] B = { 6, 5, 4, 2, 3, 1 }; int N = A.Length; Console.WriteLine(minOperations(A, B, N)); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the minimum number of // operations to delete all elements of // permutation A in order described by B function minOperations(A, B, N) { // Stores the count of operations let cnt = 0; // Stores the index of current integer // in B to be deleted let i = 0; // Loop to iterate over all values of B while (i < N) { // Stores the current index in A let j = 0; // Iterate over all values A while (j < N && i < N) { // If current integer of B and A // equal, increment the index of // the current integer of B if (B[i] == A[j]) { i++; } j++; } // As the permutation A has been // traversed completely, increment // the count of operations by 1 cnt++; } // Return Answer return cnt; } // Driver Code let A = [2, 4, 6, 1, 5, 3]; let B = [6, 5, 4, 2, 3, 1]; let N = A.length; document.write(minOperations(A, B, N)); // This code is contributed by Potta Lokesh </script> |
Output:
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
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