Minimum operations to transform given string to another by moving characters to front or end
Last Updated :
06 Apr, 2023
Given two Strings S and T of length N consisting of lowercase alphabets, which are permutations of each other, the task is to print the minimum number of operations to convert S to T. In one operation, pick any character of the string S and move it either to the start or end of the string S.
Examples:
Input: S = “abcde”, T = “edacb”
Output: 3
Explanation:
We can convert S to T in 3 moves:
1. move ‘d’ to start: “dabce”
2. move ‘e’ to start: “edabc”
3. move ‘b’ to end: “edacb”
Input: S = “dcdb”, T = “ddbc”
Output: 1
Explanation:
Move ‘c’ to end
Naive Approach: The naive approach is to try all possibilities of swapping a character. One can put some character to the front, to the end, or can leave it in the same position. The above three operations can be solved using recursion and print the minimum number of steps required after all the steps.
Time Complexity: O(3N), where N is the length of the given string.
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to observe that after moving the characters of the string S, the unchanged characters come together to form a contiguous substring in T. So, if we can maximize the length of this subsequence, then the count of operations to convert string S to T is:
N – length of the longest contiguous substring of T that is a subsequence of S
Therefore, to find the length of the longest contiguous substring of T that is a subsequence of string S, find the longest common subsequence of S and T. Let dp[][] stores the length of the longest contiguous substring of T that is a subsequence of string S, . Now dp[i][j] will store the length of the longest suffix of T[0, …, j] that is also a subsequence of S[0, …, i]. The recurrence relation is given by:
- If i is greater than 0, dp[i][j] = max(dp[i-1][j], dp[i][j]).
- If S[i] is equals to T[i] then, dp[i][j] = 1 + dp[i-1][j-1].
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int dp[1010][1010];
int solve(string s, string t)
{
int n = s.size();
int r = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
dp[i][j] = 0;
if (i > 0) {
dp[i][j] = max(dp[i - 1][j],
dp[i][j]);
}
if (s[i] == t[j]) {
int ans = 1;
if (i > 0 && j > 0) {
ans = 1 + dp[i - 1][j - 1];
}
dp[i][j] = max(dp[i][j], ans);
r = max(r, dp[i][j]);
}
}
}
return (n - r);
}
int main()
{
string s = "abcde";
string t = "edacb";
cout << solve(s, t);
return 0;
}
|
Java
class GFG{
static int[][] dp = new int[1010][1010];
static int solve(String s, String t)
{
int n = s.length();
int r = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
dp[i][j] = 0;
if (i > 0)
{
dp[i][j] = Math.max(dp[i - 1][j],
dp[i][j]);
}
if (s.charAt(i) == t.charAt(j))
{
int ans = 1;
if (i > 0 && j > 0)
{
ans = 1 + dp[i - 1][j - 1];
}
dp[i][j] = Math.max(dp[i][j], ans);
r = Math.max(r, dp[i][j]);
}
}
}
return (n - r);
}
public static void main(String[] args)
{
String s = "abcde";
String t = "edacb";
System.out.print(solve(s, t));
}
}
|
Python3
dp = [[0] * 1010] * 1010
def solve(s, t):
n = len(s)
r = 0
for j in range(0, n):
for i in range(0, n):
dp[i][j] = 0
if (i > 0):
dp[i][j] = max(dp[i - 1][j],
dp[i][j])
if (s[i] == t[j]):
ans = 1
if (i > 0 and j > 0):
ans = 1 + dp[i - 1][j - 1]
dp[i][j] = max(dp[i][j], ans)
r = max(r, dp[i][j])
return (n - r)
s = "abcde"
t = "edacb"
print(solve(s, t))
|
C#
using System;
class GFG{
static int[, ] dp = new int[1010, 1010];
static int solve(String s, String t)
{
int n = s.Length;
int r = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
dp[i, j] = 0;
if (i > 0)
{
dp[i, j] = Math.Max(dp[i - 1, j],
dp[i, j]);
}
if (s[i] == t[j])
{
int ans = 1;
if (i > 0 && j > 0)
{
ans = 1 + dp[i - 1, j - 1];
}
dp[i, j] = Math.Max(dp[i, j], ans);
r = Math.Max(r, dp[i, j]);
}
}
}
return (n - r);
}
public static void Main(String[] args)
{
String s = "abcde";
String t = "edacb";
Console.Write(solve(s, t));
}
}
|
Javascript
<script>
var dp = Array.from(Array(1010), ()=> Array(1010));
function solve(s, t)
{
var n = s.length;
var r = 0;
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
dp[i][j] = 0;
if (i > 0) {
dp[i][j] = Math.max(dp[i - 1][j],
dp[i][j]);
}
if (s[i] == t[j]) {
var ans = 1;
if (i > 0 && j > 0) {
ans = 1 + dp[i - 1][j - 1];
}
dp[i][j] = Math.max(dp[i][j], ans);
r = Math.max(r, dp[i][j]);
}
}
}
return (n - r);
}
var s = "abcde";
var t = "edacb";
document.write( solve(s, t));
</script>
|
Time Complexity: O(N2), where N is the length of the given string
Auxiliary Space: O(N2)
Efficient approach : Space optimization using 2 vectors
In this approach we use two vectors because in previous approach we can see that dp[i][j] is only dependent on the current row and previous row of dp.
dp[i][j] = max(dp[i – 1][j], dp[i][j]);
Implementation Steps :
- Made 2 vectors says curr and prev that use to keep track of values of current and previous row of matrix respectively.
- Now change dp[i] to curr and dp[i-1] to prev in previous approach.
- After every iteration of outer loop store all values of curr to prev and move to the next iterations
Implementation :
C++
#include <bits/stdc++.h>
using namespace std;
int solve(string s, string t)
{
int n = s.size();
vector<int>prev(n+1);
vector<int>curr(n+1);
int r = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
curr[j] = 0;
if (i > 0) {
curr[j] = max(prev[j], curr[j]);
}
if (s[i] == t[j]) {
int ans = 1;
if (i > 0 && j > 0) {
ans = 1 + prev[j - 1];
}
curr[j] = max(curr[j], ans);
r = max(r, curr[j]);
}
}
prev = curr;
}
return (n - r);
}
int main()
{
string s = "abcde";
string t = "edacb";
cout << solve(s, t);
return 0;
}
|
Java
import java.util.*;
public class Main {
static int solve(String s, String t) {
int n = s.length();
int[] prev = new int[n+1];
int[] curr = new int[n+1];
int r = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
curr[j] = 0;
if (i > 0) {
curr[j] = Math.max(prev[j], curr[j]);
}
if (s.charAt(i) == t.charAt(j)) {
int ans = 1;
if (i > 0 && j > 0) {
ans = 1 + prev[j - 1];
}
curr[j] = Math.max(curr[j], ans);
r = Math.max(r, curr[j]);
}
}
prev = curr.clone();
}
return (n - r);
}
public static void main(String[] args) {
String s = "abcde";
String t = "edacb";
System.out.println(solve(s, t));
}
}
|
Python3
def solve(s, t):
n = len(s)
prev = [0] * (n + 1)
curr = [0] * (n + 1)
r = 0
for i in range(n):
for j in range(n):
curr[j] = 0
if i > 0:
curr[j] = max(prev[j], curr[j])
if s[i] == t[j]:
ans = 1
if i > 0 and j > 0:
ans = 1 + prev[j - 1]
curr[j] = max(curr[j], ans)
r = max(r, curr[j])
prev = curr[:]
return (n - r)
if __name__ == '__main__':
s = "abcde"
t = "edacb"
print(solve(s, t))
|
C#
using System;
public class MainClass {
static int Solve(string s, string t)
{
int n = s.Length;
int[] prev = new int[n + 1];
int[] curr = new int[n + 1];
int r = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
curr[j] = 0;
if (i > 0) {
curr[j] = Math.Max(prev[j], curr[j]);
}
if (s[i] == t[j]) {
int ans = 1;
if (i > 0 && j > 0) {
ans = 1 + prev[j - 1];
}
curr[j] = Math.Max(curr[j], ans);
r = Math.Max(r, curr[j]);
}
}
prev = (int[])curr.Clone();
}
return (n - r);
}
public static void Main(string[] args)
{
string s = "abcde";
string t = "edacb";
Console.WriteLine(Solve(s, t));
}
}
|
Javascript
function solve(s, t) {
const n = s.length;
let prev = new Array(n + 1).fill(0);
let curr = new Array(n + 1).fill(0);
let r = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
curr[j] = 0;
if (i > 0) {
curr[j] = Math.max(prev[j], curr[j]);
}
if (s.charAt(i) == t.charAt(j)) {
let ans = 1;
if (i > 0 && j > 0) {
ans = 1 + prev[j - 1];
}
curr[j] = Math.max(curr[j], ans);
r = Math.max(r, curr[j]);
}
}
prev = curr.slice();
}
return n - r;
}
const s = "abcde";
const t = "edacb";
console.log(solve(s, t));
|
Time Complexity: O(N2), where N is the length of the given string
Auxiliary Space: O(N) only use 1d vector not 2d matrix to store values.
Note: The above naive approach is efficient for smaller strings whereas, the above efficient approach is efficient for larger strings.
Similar Reads
Minimum cost to convert one given string to another using swap, insert or delete operations
Given two strings A and B of length N and M respectively, the task is to find the minimum cost to convert string A to B using the following operations: A character of string A can be swapped from another character of the same string. Cost = 0.A character can be deleted from string B or can be insert
6 min read
Transform One String to Another using Minimum Number of Given Operation
Given two strings A and B, the task is to convert A to B if possible. The only operation allowed is to put any character from A and insert it at front. Find if it's possible to convert the string. If yes, then output minimum no. of operations required for transformation. Examples: Input: A = "ABD",
15+ min read
Minimum number of given operations required to convert a string to another string
Given two strings S and T of equal length. Both strings contain only the characters '0' and '1'. The task is to find the minimum number of operations to convert string S to T. There are 2 types of operations allowed on string S: Swap any two characters of the string.Replace a '0' with a '1' or vice
15 min read
Minimum number of Appends of X or Y characters from the end to the front required to obtain given string
Given a string S and two positive integers X and Y, the task is to find the minimum number of operations required to obtain the original string. In each operation, append X or Y characters from the end of the string to the front of the string respectively in each operation. Examples: Input: S = "Gee
9 min read
Check if it is possible to transform one string to another
Given two strings s1 and s2(all letters in uppercase). Check if it is possible to convert s1 to s2 by performing following operations. Make some lowercase letters uppercase. Delete all the lowercase letters. Examples: Input : s1 = daBcd s2 = ABC Output : yes Explanation : daBcd -> dABCd -> ABC
7 min read
Minimize operations to make one string contain only characters from other string
Given two strings S1 and S2 containing only lower case English alphabets, the task is to minimize the number of operations required to make S1 contain only characters from S2 where in each operation any character of S1 can be converted to any other letter and the cost of the operation will be differ
9 min read
Transform string A into B by deleting characters from ends and reinserting at any position
Given two strings A and B that are anagrams of each other, the task is to convert A to B if possible in minimum number of operations. An operation is defined as removing either the first or the last character in A and inserting it back anywhere in the string. Examples: Input: A = "edacb", B = "abcde
13 min read
Count characters to be shifted from the start or end of a string to obtain another string
Given two strings A and B where string A is an anagram of string B. In one operation, remove the first or the last character of the string A and insert at any position in A. The task is to find the minimum number of such operations required to be performed to convert string A into string B. Examples
6 min read
Minimum cost to convert str1 to str2 with the given operations
Given two strings of equal lengths str1 and str2 consist of characters 'a' and 'b' only. The following operations can be performed on str1: Any character can be changed from 'a' to 'b' or from 'b' to 'a' with 1 unit cost.Any two characters str1[i] and str1[j] can be swapped with cost |i - j|. The ta
6 min read
Maximum point to convert string S to T by swapping adjacent characters
Given two strings S and T of length N, the task is to find the maximum points gained in converting the string S into string T by performing the following operation on string S: In one move, select an index i (1 ≤ i ≤ N−1), get (Si − Si + 1) points, and then swap Si and Si + 1Examples: Input: S = "32
9 min read