Minimize steps to form string S from any random string of length K using a fixed length subsequences

Last Updated : 11 Oct, 2022

Given a string S consisting of N characters and a positive integer K, the task is to find the minimum number of operations required to generate the string S from a random string temp of size K and inserting the subsequence of any fixed length from the random string in the random string. If it is impossible to generate the string S, then print "-1".

Examples:

Input: S = "toffee", K = 4
Output: 2 tofe
Explanation:
Consider the random string temp as "tofe" which is of length K(= 4) and perform the following steps:
Operation 1: Take a subsequence of length 1 from string temp as 'f' and after inserting the subsequence in the string "tofe" modifies the string to "toffe".
Operation 2: Take a subsequence of length 1 from string temp as 'e' and after inserting the subsequence in the string "toffe" modifies the string to "toffee".
Therefore, the minimum number of operations required is 2.

Input: S = "book", K = 2
Output: -1

Approach: This problem can be solved by using Greedy Approach and Binary Search. Follow the steps below to solve this problem:

Initialize an array, say amounts[] that stores the frequency of each character of string S.
Iterate in the range [0, N - 1] using the variable i and increment the frequency of the current character by 1 in the array amounts[].
Find count unique characters in the string S and store it in a variable, say, count.
If the count is greater than N, then return -1.
Now, perform the Binary Search to find the resultant string and perform the following steps:
Initialize two variables high as 10⁵ and low as 0.
Iterate until the value of (high - low) is greater than 1 and perform the following steps:
- Update the value of total as 0 and mid as (high + low) / 2.
- Iterate in the range [0, 25] using the variable i and if the value of amounts[i] is greater than 0, then increment the total by (amounts[i] - 1)/mid + 1.
- If the total is less than equal to N, then update the value of high as mid. Otherwise, update the value of low as mid.
After completing the above steps, print the value of high and iterate in the range[0, 25], and print the resultant string.

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 string required to generate // the original string void findString(string S, int K) {     // Stores the frequency of each     // character of String S     int amounts[26];      // Iterate over the range [0, 25]     for (int i = 0; i < 26; i++) {         amounts[i] = 0;     }      // Stores the frequency of each     // character of String S     for (int i = 0; i < S.length(); i++) {         amounts[int(S[i]) - 97]++;     }      int count = 0;      // Count unique characters in S     for (int i = 0; i < 26; i++) {         if (amounts[i] > 0)             count++;     }      // If unique characters is greater     // then N, then return -1     if (count > K) {         cout << "-1";     }      // Otherwise     else {          string ans = "";         int high = 100001;         int low = 0;         int mid, total;          // Perform Binary Search         while ((high - low) > 1) {              total = 0;              // Find the value of mid             mid = (high + low) / 2;              // Iterate over the range             // [0, 26]             for (int i = 0; i < 26; i++) {                  // If the amount[i] is                 // greater than 0                 if (amounts[i] > 0) {                     total += (amounts[i] - 1)                                  / mid                              + 1;                 }             }              // Update the ranges             if (total <= K) {                 high = mid;             }             else {                 low = mid;             }         }          cout << high << " ";          total = 0;          // Find the resultant string         for (int i = 0; i < 26; i++) {              if (amounts[i] > 0) {                 total += (amounts[i] - 1)                              / high                          + 1;                  for (int j = 0;                      j < ((amounts[i] - 1)                               / high                           + 1);                      j++) {                      // Generate the subsequence                     ans += char(i + 97);                 }             }         }          // If the length of resultant         // string is less than N than         // add a character 'a'         for (int i = total; i < K; i++) {             ans += 'a';         }          reverse(ans.begin(), ans.end());          // Print the string         cout << ans;     } }  // Driver Code int main() {     string S = "toffee";     int K = 4;     findString(S, K);      return 0; }

Java

// Java code for above approach import java.util.*;  class GFG{  // Function to find the minimum number // of string required to generate // the original string static void findString(String S, int N) {        // Stores the frequency of each     // character of string S     int[] amounts = new int[26];      // Iterate over the range [0, 25]     for (int i = 0; i < 26; i++) {         amounts[i] = 0;     }      // Stores the frequency of each     // character of string S     for (int i = 0; i < S.length(); i++) {         amounts[(int)(S.charAt(i) - 97)]++;     }      int count = 0;      // Count unique characters in S     for (int i = 0; i < 26; i++) {         if (amounts[i] > 0)             count++;     }      // If unique characters is greater     // then N, then return -1     if (count > N) {         System.out.print("-1");     }      // Otherwise     else {          String ans = "";         int high = 100001;         int low = 0;         int mid, total;          // Perform Binary Search         while ((high - low) > 1) {              total = 0;              // Find the value of mid             mid = (high + low) / 2;              // Iterate over the range             // [0, 26]             for (int i = 0; i < 26; i++) {                  // If the amount[i] is                 // greater than 0                 if (amounts[i] > 0) {                     total += (amounts[i] - 1)                                  / mid                              + 1;                 }             }              // Update the ranges             if (total <= N) {                 high = mid;             }             else {                 low = mid;             }         }          System.out.print(high + " ");          total = 0;          // Find the resultant string         for (int i = 0; i < 26; i++) {              if (amounts[i] > 0) {                 total += (amounts[i] - 1)                              / high                          + 1;                  for (int j = 0;                      j < ((amounts[i] - 1)                               / high                           + 1);                      j++) {                      // Generate the subsequence                     ans += (char)(i + 97);                 }             }         }          // If the length of resultant         // string is less than N than         // add a character 'a'         for (int i = total; i < N; i++) {             ans += 'a';         }                  String reverse = "";          int Len = ans.length() - 1;               while(Len >= 0)               {                   reverse = reverse + ans.charAt(Len);                   Len--;               }           // Print the string         System.out.print(reverse);     } }  // Driver Code public static void main(String[] args) {     String S = "toffee";     int K = 4;     findString(S, K); } }  // This code is contributed by target_2.

Python3

# Python3 program for the above approach   # Function to find the minimum number  # of string required to generate  # the original string  def findString(S, N):          # Stores the frequency of each      # character of String S      amounts = [0] * 26          # Stores the frequency of each      # character of String S      for i in range(len(S)):         amounts[ord(S[i]) - 97] += 1              count = 0          # Count unique characters in S      for i in range(26):         if amounts[i] > 0:             count += 1                  # If unique characters is greater      # then N, then return -1      if count > N:         print("-1")              # Otherwise      else:         ans = ""         high = 100001         low = 0                  # Perform Binary Search          while (high - low) > 1:             total = 0                          # Find the value of mid              mid = (high + low) // 2                          # Iterate over the range              # [0, 26]              for i in range(26):                                  # If the amount[i] is                  # greater than 0                  if amounts[i] > 0:                     total += (amounts[i] - 1) // mid + 1                                  # Update the ranges              if total <= N:                 high = mid             else:                 low = mid                          print(high, end = " ")         total = 0                  # Find the resultant string          for i in range(26):             if amounts[i] > 0:                 total += (amounts[i] - 1) // high + 1                 for j in range((amounts[i] - 1) // high + 1):                     ans += chr(i + 97)                  # If the length of resultant          # string is less than N than          # add a character 'a'                      for i in range(total, N):             ans += 'a'                      ans = ans[::-1]                  # Print the string          print(ans)  # Driver code S = "toffee" K = 4  findString(S, K)  # This code is contributed by Parth Manchanda

// C# program for the above approach using System;  class GFG{      // Function to find the minimum number // of string required to generate // the original string static void findString(string S, int N) {        // Stores the frequency of each     // character of string S     int[] amounts = new int[26];      // Iterate over the range [0, 25]     for (int i = 0; i < 26; i++) {         amounts[i] = 0;     }      // Stores the frequency of each     // character of string S     for (int i = 0; i < S.Length; i++) {         amounts[(int)(S[i] - 97)]++;     }      int count = 0;      // Count unique characters in S     for (int i = 0; i < 26; i++) {         if (amounts[i] > 0)             count++;     }      // If unique characters is greater     // then N, then return -1     if (count > N) {         Console.Write("-1");     }      // Otherwise     else {          string ans = "";         int high = 100001;         int low = 0;         int mid, total;          // Perform Binary Search         while ((high - low) > 1) {              total = 0;              // Find the value of mid             mid = (high + low) / 2;              // Iterate over the range             // [0, 26]             for (int i = 0; i < 26; i++) {                  // If the amount[i] is                 // greater than 0                 if (amounts[i] > 0) {                     total += (amounts[i] - 1)                                  / mid                              + 1;                 }             }              // Update the ranges             if (total <= N) {                 high = mid;             }             else {                 low = mid;             }         }          Console.Write(high + " ");          total = 0;          // Find the resultant string         for (int i = 0; i < 26; i++) {              if (amounts[i] > 0) {                 total += (amounts[i] - 1)                              / high                          + 1;                  for (int j = 0;                      j < ((amounts[i] - 1)                               / high                           + 1);                      j++) {                      // Generate the subsequence                     ans += (char)(i + 97);                 }             }         }          // If the length of resultant         // string is less than N than         // add a character 'a'         for (int i = total; i < N; i++) {             ans += 'a';         }                  string reverse = "";          int Len = ans.Length - 1;               while(Len >= 0)               {                   reverse = reverse + ans[Len];                   Len--;               }           // Print the string         Console.Write(reverse);     } }  // Driver Code public static void Main() {     string S = "toffee";     int K = 4;     findString(S, K); } }  // This code is contributed by splevel62.

JavaScript

<script> // Javascript program for the above approach  // Function to find the minimum number // of string required to generate // the original string function findString(S, N) {    // Stores the frequency of each   // character of String S   let amounts = new Array(26);    // Iterate over the range [0, 25]   for (let i = 0; i < 26; i++) {     amounts[i] = 0;   }    // Stores the frequency of each   // character of String S   for (let i = 0; i < S.length; i++) {     amounts[S[i].charCodeAt(0) - 97]++;   }    let count = 0;    // Count unique characters in S   for (let i = 0; i < 26; i++) {     if (amounts[i] > 0) count++;   }    // If unique characters is greater   // then N, then return -1   if (count > N) {     document.write("-1");   }    // Otherwise   else {     let ans = "";     let high = 100001;     let low = 0;     let mid, total;      // Perform Binary Search     while (high - low > 1) {       total = 0;        // Find the value of mid       mid = Math.floor((high + low) / 2);        // Iterate over the range       // [0, 26]       for (let i = 0; i < 26; i++) {         // If the amount[i] is         // greater than 0         if (amounts[i] > 0) {           total += Math.floor((amounts[i] - 1) / mid + 1);         }       }        // Update the ranges       if (total <= N) {         high = mid;       } else {         low = mid;       }     }      document.write(high + " ");      total = 0;      // Find the resultant string     for (let i = 0; i < 26; i++) {       if (amounts[i] > 0) {         total += Math.floor((amounts[i] - 1) / high + 1);          for (let j = 0; j < Math.floor((amounts[i] - 1) / high + 1); j++) {           // Generate the subsequence           ans += String.fromCharCode(i + 97);         }       }     }      // If the length of resultant     // string is less than N than     // add a character 'a'     for (let i = total; i < N; i++) {       ans += "a";     }      ans =  ans.split("").reverse().join("");      // Print the string     document.write(ans);   } }  // Driver Code  let S = "toffee"; let K = 4; findString(S, K);  // This code is contributed by _saurabh_jaiswal. </script>

Output:

2 tofe

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(K), for storing a string of length K.

Minimize length of a string by removing occurrences of another string from it as a substring

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