Minimum number of given operations required to make two strings equal
Given two strings A and B, both strings contain characters a and b and are of equal lengths. There is one _ (empty space) in both the strings. The task is to convert first string into second string by doing the minimum number of the following operations:
- If _ is at position i then _ can be swapped with a character at position i+1 or i-1.
- If characters at positions i+1 and i+2 are different then _ can be swapped with a character at position i+1 or i+2.
- Similarly, if characters at positions i-1 and i-2 are different then _ can be swapped with a character at position i-1 or i-2.
Examples:
Input: A = “aba_a”, B = “_baaa”
Output: 2
Move 1 : A = “ab_aa” (Swapped A[2] with A[3])
Move 2 : A = “_baaa” (Swapped A[0] with A[2])
Input: A = “a_b”, B = “ab_”
Output: 1
Source: Directi Interview Set 7
Approach:
- Apply a simple Breadth First Search over the string and an element of the queue used for BFS will contain the pair str, pos where pos is the position of _ in the string str.
- Also maintain a map vis which will store the string as key and the minimum moves to get to the string as value.
- For every string str from the queue, generate a new string tmp based on the four conditions given and update the vis map as vis[tmp] = vis[str] + 1.
- Repeat the above steps until the queue is empty or the required string is generated i.e. tmp == B
- If the required string is generated then return vis[str] + 1 which is the minimum number of operations required to change A to B.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to return minimum number of // operations to convert string A to B int minOperations(string s, string f) { unordered_map<string, int > vis; int n; n = s.length(); int pos = 0; for ( int i = 0; i < s.length(); i++) { if (s[i] == '_' ) { // store the position of '_' pos = i; break ; } } // to store the generated string at every // move and the position of '_' within it queue<pair<string, int > > q; q.push({ s, pos }); // vis will store the minimum operations // to reach that particular string vis[s] = 0; while (!q.empty()) { string ss = q.front().first; int pp = q.front().second; // minimum moves to reach the string ss int dist = vis[ss]; q.pop(); // try all 4 possible operations // if '_' can be swapped with // the character on it's left if (pp > 0) { // swap with the left character swap(ss[pp], ss[pp - 1]); // if the string is generated // for the first time if (!vis.count(ss)) { // if generated string is // the required string if (ss == f) { return dist + 1; break ; } // update the distance for the // currently generated string vis[ss] = dist + 1; q.push({ ss, pp - 1 }); } // restore the string before it was // swapped to check other cases swap(ss[pp], ss[pp - 1]); } // swap '_' with the character // on it's right this time if (pp < n - 1) { swap(ss[pp], ss[pp + 1]); if (!vis.count(ss)) { if (ss == f) { return dist + 1; break ; } vis[ss] = dist + 1; q.push({ ss, pp + 1 }); } swap(ss[pp], ss[pp + 1]); } // if '_' can be swapped // with the character 'i+2' if (pp > 1 && ss[pp - 1] != ss[pp - 2]) { swap(ss[pp], ss[pp - 2]); if (!vis.count(ss)) { if (ss == f) { return dist + 1; break ; } vis[ss] = dist + 1; q.push({ ss, pp - 2 }); } swap(ss[pp], ss[pp - 2]); } // if '_' can be swapped // with the character at 'i+2' if (pp < n - 2 && ss[pp + 1] != ss[pp + 2]) { swap(ss[pp], ss[pp + 2]); if (!vis.count(ss)) { if (ss == f) { return dist + 1; break ; } vis[ss] = dist + 1; q.push({ ss, pp + 2 }); } swap(ss[pp], ss[pp + 2]); } } } // Driver code int main() { string A = "aba_a" ; string B = "_baaa" ; cout << minOperations(A, B); return 0; } |
2
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