# Least number of manipulations needed to ensure two strings have identical characters

Given two strings return the value of least number of manipulations needed to ensure both strings have identical characters, i.e., both string become anagram of each other.

**Examples:**

Input : s1 = "aab" s2 = "aba" Output : 2 Explanation : string 1 contains 2 a's and 1 b, also string 2 contains same characters Input : s1 = "abc" s2 = "cdd" Output : 2 Explanation : string 1 contains 1 a, 1 b, 1 c while string 2 contains 1 c and 2 d's so there are 2 different characters

**Question Source :** Yatra.com Interview Experience | Set 7

The idea is to create a extra count array for both the strings separately and then count the difference in characters.

## C++

`// CPP program to count least number ` `// of manipulations to have two strings ` `// set of same characters ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `const` `int` `MAX_CHAR = 26; ` ` ` `// return the count of manipulations ` `// required ` `int` `leastCount(string s1, string s2, ` `int` `n) ` `{ ` ` ` `int` `count1[MAX_CHAR] = { 0 }; ` ` ` `int` `count2[MAX_CHAR] = { 0 }; ` ` ` ` ` `// count the number of different ` ` ` `// characters in both strings ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `count1[s1[i] - ` `'a'` `] += 1; ` ` ` `count2[s2[i] - ` `'a'` `] += 1; ` ` ` `} ` ` ` ` ` `// check the difference in characters ` ` ` `// by comparing count arrays ` ` ` `int` `res = 0; ` ` ` `for` `(` `int` `i = 0; i < MAX_CHAR; i++) { ` ` ` `if` `(count1[i] != 0) { ` ` ` `res += ` `abs` `(count1[i] - count2[i]); ` ` ` `} ` ` ` `} ` ` ` `return` `res; ` `} ` ` ` `// driver program ` `int` `main() ` `{ ` ` ` `string s1 = ` `"abc"` `; ` ` ` `string s2 = ` `"cdd"` `; ` ` ` `int` `len = s1.length(); ` ` ` `int` `res = leastCount(s1, s2, len); ` ` ` `cout << res << endl; ` ` ` `return` `0; ` `} ` |

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## Java

`// java program to count least number ` `// of manipulations to have two ` `// strings set of same characters ` `import` `java.io.*; ` ` ` `public` `class` `GFG { ` ` ` ` ` `static` `int` `MAX_CHAR = ` `26` `; ` ` ` ` ` `// return the count of manipulations ` ` ` `// required ` ` ` `static` `int` `leastCount(String s1, ` ` ` `String s2, ` `int` `n) ` ` ` `{ ` ` ` ` ` `int` `[] count1 = ` `new` `int` `[MAX_CHAR]; ` ` ` `int` `[] count2 = ` `new` `int` `[MAX_CHAR]; ` ` ` ` ` `// count the number of different ` ` ` `// characters in both strings ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `{ ` ` ` `count1[s1.charAt(i) - ` `'a'` `] += ` `1` `; ` ` ` `count2[s2.charAt(i) - ` `'a'` `] += ` `1` `; ` ` ` `} ` ` ` ` ` `// check the difference in characters ` ` ` `// by comparing count arrays ` ` ` `int` `res = ` `0` `; ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < MAX_CHAR; i++) ` ` ` `{ ` ` ` `if` `(count1[i] != ` `0` `) { ` ` ` `res += Math.abs(count1[i] ` ` ` `- count2[i]); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// driver program ` ` ` `static` `public` `void` `main(String[] args) ` ` ` `{ ` ` ` `String s1 = ` `"abc"` `; ` ` ` `String s2 = ` `"cdd"` `; ` ` ` `int` `len = s1.length(); ` ` ` `int` `res = leastCount(s1, s2, len); ` ` ` ` ` `System.out.println(res); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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## Python3

`# Python3 program to count least number ` `# of manipulations to have two strings ` `# set of same characters ` `MAX_CHAR ` `=` `26` ` ` `# return the count of manipulations ` `# required ` `def` `leastCount(s1, s2, n): ` ` ` ` ` `count1 ` `=` `[` `0` `] ` `*` `MAX_CHAR ` ` ` `count2 ` `=` `[` `0` `] ` `*` `MAX_CHAR ` ` ` ` ` `# count the number of different ` ` ` `# characters in both strings ` ` ` `for` `i ` `in` `range` `( n): ` ` ` `count1[` `ord` `(s1[i]) ` `-` `ord` `(` `'a'` `)] ` `+` `=` `1` ` ` `count2[` `ord` `(s2[i]) ` `-` `ord` `(` `'a'` `)] ` `+` `=` `1` ` ` ` ` `# check the difference in characters ` ` ` `# by comparing count arrays ` ` ` `res ` `=` `0` ` ` `for` `i ` `in` `range` `(MAX_CHAR): ` ` ` `if` `(count1[i] !` `=` `0` `): ` ` ` `res ` `+` `=` `abs` `(count1[i] ` `-` `count2[i]) ` ` ` ` ` `return` `res ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `s1 ` `=` `"abc"` ` ` `s2 ` `=` `"cdd"` ` ` `l ` `=` `len` `(s1) ` ` ` `res ` `=` `leastCount(s1, s2, l) ` ` ` `print` `(res) ` ` ` `# This code is contributed by ita_c ` |

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## C#

`// C# program to count least number ` `// of manipulations to have two strings ` `// set of same characters ` `using` `System; ` ` ` `public` `class` `GFG { ` ` ` ` ` `static` `int` `MAX_CHAR = 26; ` ` ` ` ` `// return the count of manipulations ` ` ` `// required ` ` ` `static` `int` `leastCount(` `string` `s1, ` ` ` `string` `s2, ` `int` `n) ` ` ` `{ ` ` ` ` ` `int` `[] count1 = ` `new` `int` `[MAX_CHAR]; ` ` ` `int` `[] count2 = ` `new` `int` `[MAX_CHAR]; ` ` ` ` ` `// count the number of different ` ` ` `// characters in both strings ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `count1[s1[i] - ` `'a'` `] += 1; ` ` ` `count2[s2[i] - ` `'a'` `] += 1; ` ` ` `} ` ` ` ` ` `// check the difference in characters ` ` ` `// by comparing count arrays ` ` ` `int` `res = 0; ` ` ` `for` `(` `int` `i = 0; i < MAX_CHAR; i++) ` ` ` `{ ` ` ` `if` `(count1[i] != 0) { ` ` ` `res += Math.Abs(count1[i] ` ` ` `- count2[i]); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// driver program ` ` ` `static` `public` `void` `Main() ` ` ` `{ ` ` ` `string` `s1 = ` `"abc"` `; ` ` ` `string` `s2 = ` `"cdd"` `; ` ` ` `int` `len = s1.Length; ` ` ` `int` `res = leastCount(s1, s2, len); ` ` ` `Console.WriteLine(res); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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**Output:**

2

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