Given two strings, find if we can make first string from second by deleting some characters from second and rearranging remaining characters.
Examples:
Input : s1 = ABHISHEKsinGH, s2 = gfhfBHkooIHnfndSHEKsiAnG
Output : PossibleInput : s1 = Hello, s2 = dnaKfhelddf
Output : Not PossibleInput : s1 = neveropen, s2 = rteksfoGrdsskGeggehes
Output : Possible
We basically need to find if one string contains characters which are subset of characters in second string. First we count occurrences of all characters in second string. Then we traverse through first string and reduce count of every character that is present in first. If at any moment, count becomes less than 0, we return false. If all counts remain greater than or equal to 0, we return true.
Implementation:
C++
// CPP program to find if it is possible to// make a string from characters present in// other string.#include <bits/stdc++.h>using namespace std;const int MAX_CHAR = 256;// Returns true if it is possible to make// s1 from characters present in s2.bool isPossible(string& s1, string& s2){ // Count occurrences of all characters // present in s2.. int count[MAX_CHAR] = { 0 }; for (int i = 0; i < s2.length(); i++) count[s2[i]]++; // For every character, present in s1, // reduce its count if count is more // than 0. for (int i = 0; i < s1.length(); i++) { if (count[s1[i]] == 0) return false; count[s1[i]]--; } return true;}// Driver codeint main(){ string s1 = "neveropen", s2 = "rteksfoGrdsskGeggehes"; if (isPossible(s1, s2)) cout << "Possible"; else cout << "Not Possible\n"; return 0;} |
Java
// Java program to find if it is possible to// make a string from characters present in// other string.class StringfromCharacters{ static final int MAX_CHAR = 256; // Returns true if it is possible to make // s1 from characters present in s2. static boolean isPossible(String s1, String s2) { // Count occurrences of all characters // present in s2.. int count[] = new int[MAX_CHAR]; for (int i = 0; i < s2.length(); i++) count[(int)s2.charAt(i)]++; // For every character, present in s1, // reduce its count if count is more // than 0. for (int i = 0; i < s1.length(); i++) { if (count[(int)s1.charAt(i)] == 0) return false; count[(int)s1.charAt(i)]--; } return true; } // Driver code public static void main(String args[]) { String s1 = "neveropen", s2 = "rteksfoGrdsskGeggehes"; if (isPossible(s1, s2)) System.out.println("Possible"); else System.out.println("Not Possible"); }}/* This code is contributed by Danish Kaleem */ |
Python 3
# Python 3 program to find if it is possible # to make a string from characters present # in other string.MAX_CHAR = 256# Returns true if it is possible to make# s1 from characters present in s2.def isPossible(s1, s2): # Count occurrences of all characters # present in s2.. count = [0] * MAX_CHAR for i in range(len(s2)): count[ord(s2[i])] += 1 # For every character, present in s1, # reduce its count if count is more # than 0. for i in range(len(s1)): if (count[ord(s1[i])] == 0): return False count[ord(s1[i])] -= 1 return True# Driver codeif __name__ == "__main__": s1 = "neveropen" s2 = "rteksfoGrdsskGeggehes" if (isPossible(s1, s2)): print("Possible") else: print("Not Possible")# This code is contributed by ita_c |
C#
// C# program to find if it is possible to// make a string from characters present// in other string.using System;class GFG { static int MAX_CHAR = 256; // Returns true if it is possible to // make s1 from characters present // in s2. static bool isPossible(String s1, String s2) { // Count occurrences of all characters // present in s2.. int []count = new int[MAX_CHAR]; for (int i = 0; i < s2.Length; i++) count[(int)s2[i]]++; // For every character, present in s1, // reduce its count if count is more // than 0. for (int i = 0; i < s1.Length; i++) { if (count[(int)s1[i]] == 0) return false; count[(int)s1[i]]--; } return true; } // Driver code public static void Main() { string s1 = "neveropen", s2 = "rteksfoGrdsskGeggehes"; if (isPossible(s1, s2)) Console.WriteLine("Possible"); else Console.WriteLine("Not Possible"); }}// This code is contributed by vt_m. |
Javascript
<script>// Javascript program to find if it is possible to// make a string from characters present in// other string. let MAX_CHAR = 256; // Returns true if it is possible to make // s1 from characters present in s2. function isPossible(s1, s2) { // Count occurrences of all characters // present in s2.. let count = new Array(MAX_CHAR); for(let i = 0; i < count.length; i++) { count[i] = 0; } for (let i = 0; i < s2.length; i++) count[s2[i].charCodeAt(0)]++; // For every character, present in s1, // reduce its count if count is more // than 0. for (let i = 0; i < s1.length; i++) { if (count[s1[i].charCodeAt(0)] == 0) return false; count[s1[i].charCodeAt(0)]--; } return true; } // Driver code let s1 = "neveropen", s2 = "rteksfoGrdsskGeggehes"; if (isPossible(s1, s2)) document.write("Possible"); else document.write("Not Possible"); // This code is contributed by avanitrachhadiya2155</script> |
Output
Possible
Time complexity is: O(n)
Auxiliary Space: O(256)
This article is contributed by Prabhat kumar singh. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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