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

• Last Updated : 10 Aug, 2021

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”, N = 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”, N = 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 ``using` `namespace` `std;` `// Function to find the minimum number``// of string required to generate``// the original string``void` `findString(string S, ``int` `N)``{``    ``// Stores the frequency of each``    ``// character of String S``    ``int` `amounts;` `    ``// 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) {``        ``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 <= N) {``                ``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 < N; 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#

 `// 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``;` `    ``// 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

 ``
Output:
`2 tofe`

Time Complexity: O(N)
Auxiliary Space: O(1)

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