Given two strings ‘str1’ and ‘str2’ of size m and n respectively. The task is to remove/delete and insert the minimum number of characters from/in str1 to transform it into str2. It could be possible that the same character needs to be removed/deleted from one point of str1 and inserted to some another point.
Example 1:
Input : str1 = "heap", str2 = "pea" Output : Minimum Deletion = 2 and Minimum Insertion = 1 Explanation: p and h deleted from heap Then, p is inserted at the beginning One thing to note, though p was required yet it was removed/deleted first from its position and then it is inserted to some other position. Thus, p contributes one to the deletion_count and one to the insertion_count.
Example 2:
Input : str1 = "geeksforgeeks", str2 = "geeks" Output : Minimum Deletion = 8 Minimum Insertion = 0
Simple Approach:
A simple approach is to consider all subsequences of str1 and for each subsequence calculate minimum deletions and insertions so as to transform it into str2. A very complex method and the time complexity of this solution is exponential.
Efficient Approach:
An efficient approach uses the concept of finding the length of the longest common subsequence of the given two sequences.
Algorithm:
- str1 and str2 be the given strings.
- m and n be their lengths respectively.
- len be the length of the longest common subsequence of str1 and str2
- minimum number of deletions minDel = m – len
- minimum number of Insertions minInsert = n – len
Below is the implementation of the above code:
C++
// Dynamic Programming C++ implementation to find // minimum number of deletions and insertions #include <bits/stdc++.h> using namespace std; // Returns length of length common subsequence // for str1[0..m-1], str2[0..n-1] int lcs(string str1, string str2, int m, int n) { int L[m + 1][n + 1]; int i, j; // Following steps build L[m+1][n+1] in bottom // up fashion. Note that L[i][j] contains // length of LCS of str1[0..i-1] and str2[0..j-1] for (i = 0; i <= m; i++) { for (j = 0; j <= n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if (str1.at(i - 1) == str2.at(j - 1)) L[i][j] = L[i - 1][j - 1] + 1; else L[i][j] = max(L[i - 1][j], L[i][j - 1]); } } // L[m][n] contains length of LCS // for X[0..n-1] and Y[0..m-1] return L[m][n]; } // function to find minimum number // of deletions and insertions void printMinDelAndInsert(string str1, string str2) { int m = str1.size(); int n = str2.size(); int len = lcs(str1, str2, m, n); cout << "Minimum number of deletions = " << (m - len) << endl; cout << "Minimum number of insertions = " << (n - len) << endl; } // Driver Code int main() { string str1 = "heap" ; string str2 = "pea" ; // Function Call printMinDelAndInsert(str1, str2); return 0; } |
Java
// Dynamic Programming Java implementation // to find minimum number of deletions and // insertions import java.io.*; class GFG { // Returns length of length common // subsequence for str1[0..m-1], // str2[0..n-1] static int lcs(String str1, String str2, int m, int n) { int L[][] = new int [m + 1 ][n + 1 ]; int i, j; // Following steps build L[m+1][n+1] in // bottom up fashion. Note that L[i][j] // contains length of LCS of str1[0..i-1] // and str2[0..j-1] for (i = 0 ; i <= m; i++) { for (j = 0 ; j <= n; j++) { if (i == 0 || j == 0 ) L[i][j] = 0 ; else if (str1.charAt(i - 1 ) == str2.charAt(j - 1 )) L[i][j] = L[i - 1 ][j - 1 ] + 1 ; else L[i][j] = Math.max(L[i - 1 ][j], L[i][j - 1 ]); } } // L[m][n] contains length of LCS // for X[0..n-1] and Y[0..m-1] return L[m][n]; } // function to find minimum number // of deletions and insertions static void printMinDelAndInsert(String str1, String str2) { int m = str1.length(); int n = str2.length(); int len = lcs(str1, str2, m, n); System.out.println( "Minimum number of " + "deletions = " ); System.out.println(m - len); System.out.println( "Minimum number of " + "insertions = " ); System.out.println(n - len); } // Driver code public static void main(String[] args) { String str1 = new String( "heap" ); String str2 = new String( "pea" ); // Function Call printMinDelAndInsert(str1, str2); } } // This code is contributed by Prerna Saini |
Python3
# Dynamic Programming Python3 # implementation to find minimum # number of deletions and insertions # Returns length of length # common subsequence for # str1[0..m-1], str2[0..n-1] def lcs(str1, str2, m, n): L = [[ 0 for i in range (n + 1 )] for i in range (m + 1 )] # Following steps build L[m+1][n+1] # in bottom up fashion. Note that # L[i][j] contains length of LCS # of str1[0..i-1] and str2[0..j-1] for i in range (m + 1 ): for j in range (n + 1 ): if (i = = 0 or j = = 0 ): L[i][j] = 0 elif (str1[i - 1 ] = = str2[j - 1 ]): L[i][j] = L[i - 1 ][j - 1 ] + 1 else : L[i][j] = max (L[i - 1 ][j], L[i][j - 1 ]) # L[m][n] contains length of LCS # for X[0..n-1] and Y[0..m-1] return L[m][n] # function to find minimum number # of deletions and insertions def printMinDelAndInsert(str1, str2): m = len (str1) n = len (str2) leng = lcs(str1, str2, m, n) print ( "Minimum number of deletions = " , m - leng, sep = ' ' ) print ( "Minimum number of insertions = " , n - leng, sep = ' ' ) # Driver Code str1 = "heap" str2 = "pea" # Function Call printMinDelAndInsert(str1, str2) # This code is contributed # by sahilshelangia |
C#
// Dynamic Programming C# implementation // to find minimum number of deletions and // insertions using System; class GFG { // Returns length of length common // subsequence for str1[0..m-1], // str2[0..n-1] static int lcs( string str1, string str2, int m, int n) { int [, ] L = new int [m + 1, n + 1]; int i, j; // Following steps build L[m+1][n+1] in // bottom up fashion. Note that L[i][j] // contains length of LCS of str1[0..i-1] // and str2[0..j-1] for (i = 0; i <= m; i++) { for (j = 0; j <= n; j++) { if (i == 0 || j == 0) L[i, j] = 0; else if (str1[i - 1] == str2[j - 1]) L[i, j] = L[i - 1, j - 1] + 1; else L[i, j] = Math.Max(L[i - 1, j], L[i, j - 1]); } } // L[m][n] contains length of LCS // for X[0..n-1] and Y[0..m-1] return L[m, n]; } // function to find minimum number // of deletions and insertions static void printMinDelAndInsert( string str1, string str2) { int m = str1.Length; int n = str2.Length; int len = lcs(str1, str2, m, n); Console.Write( "Minimum number of " + "deletions = " ); Console.WriteLine(m - len); Console.Write( "Minimum number of " + "insertions = " ); Console.Write(n - len); } // Driver code public static void Main() { string str1 = new string ( "heap" ); string str2 = new string ( "pea" ); // Function Call printMinDelAndInsert(str1, str2); } } // This code is contributed by nitin mittal. |
Minimum number of deletions = 2 Minimum number of insertions = 1
Time Complexity: O(m * n)
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