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In this article, we will discuss how to minimize the number of given operations required to align two given strings with each other. We will follow the step-by-step approach and provide C code implementation. We will also provide a sample test case to help understand the problem and solution.
Given two strings s1 and s2, we need to find the minimum number of operations required to make s1 and s2 align with each other. We can perform two operations: swap any two characters of s1, or swap any two characters of s2.
In order to solve this problem, we need to count the number of characters that do not exist in the two strings, that is, the difference in the frequency of occurrence of characters in the two strings. The minimum number of swaps required to make two strings permute each other is equal to half this count, since we can swap characters in either string to make them equal.
First, we will use two arrays to calculate the frequency of characters in two strings. We will then iterate through the two arrays and add the absolute difference between the character frequencies to a variable. This variable will store the number of characters that are not present in both strings.
After calculating the count, we return half of it as the minimum number of swaps required for the two strings to be permuted with each other.
The following is the C code implementation of the above method -
#include<bits/stdc++.h> using namespace std; int countMinSwaps(string s1, string s2) { int freq1[26] = {0}, freq2[26] = {0}, count = 0; for (char c : s1) { freq1[c - 'a']++; } for (char c : s2) { freq2[c - 'a']++; } for (int i = 0; i < 26; i++) { count += abs(freq1[i] - freq2[i]); } return count / 2; } int main() { string s1 = "hello"; string s2 = "world"; int minSwaps = countMinSwaps(s1, s2); cout << "Minimum number of swaps required: " << minSwaps << endl; return 0; }
Minimum number of swaps required: 3
Let us consider the example strings "hello" and "world" for this test case.
The frequency arrays of the two strings are as follows -
freq1 = {0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} freq2 = {0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}
We can see that the character "l" appears with frequency 2 in s1 but only 1 in s2, while the character "r" appears with frequency 1 in s2 but does not exist in s1. Therefore, the number of characters that are not present in both strings is 3.
Therefore, the minimum number of exchanges required for two strings to be arranged with each other is 1. We can swap the "l" in s1 with the "r" in s2 to get the strings "herlo" and "wolld", which are permutations of each other.
In this article, we discussed how to minimize the number of given operations required to align two given strings with each other. We followed a step-by-step approach and provided C code implementation. We also provide a sample test case to help understand the problem and solution. The problem can be solved in O(n) time complexity and O(1) space complexity.
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