Least number of manipulations needed to ensure two strings have identical characters
Last Updated :
18 Jul, 2022
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.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;
int leastCount(string s1, string s2, int n)
{
int count1[MAX_CHAR] = { 0 };
int count2[MAX_CHAR] = { 0 };
for ( int i = 0; i < n; i++) {
count1[s1[i] - 'a' ] += 1;
count2[s2[i] - 'a' ] += 1;
}
int res = 0;
for ( int i = 0; i < MAX_CHAR; i++) {
if (count1[i] != 0) {
res += abs (count1[i] - count2[i]);
}
}
return res;
}
int main()
{
string s1 = "abc" ;
string s2 = "cdd" ;
int len = s1.length();
int res = leastCount(s1, s2, len);
cout << res << endl;
return 0;
}
|
Java
import java.io.*;
public class GFG {
static int MAX_CHAR = 26 ;
static int leastCount(String s1,
String s2, int n)
{
int [] count1 = new int [MAX_CHAR];
int [] count2 = new int [MAX_CHAR];
for ( int i = 0 ; i < n; i++)
{
count1[s1.charAt(i) - 'a' ] += 1 ;
count2[s2.charAt(i) - 'a' ] += 1 ;
}
int res = 0 ;
for ( int i = 0 ; i < MAX_CHAR; i++)
{
if (count1[i] != 0 ) {
res += Math.abs(count1[i]
- count2[i]);
}
}
return res;
}
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);
}
}
|
Python3
MAX_CHAR = 26
def leastCount(s1, s2, n):
count1 = [ 0 ] * MAX_CHAR
count2 = [ 0 ] * MAX_CHAR
for i in range ( n):
count1[ ord (s1[i]) - ord ( 'a' )] + = 1
count2[ ord (s2[i]) - ord ( 'a' )] + = 1
res = 0
for i in range (MAX_CHAR):
if (count1[i] ! = 0 ):
res + = abs (count1[i] - count2[i])
return res
if __name__ = = "__main__" :
s1 = "abc"
s2 = "cdd"
l = len (s1)
res = leastCount(s1, s2, l)
print (res)
|
C#
using System;
public class GFG {
static int MAX_CHAR = 26;
static int leastCount( string s1,
string s2, int n)
{
int [] count1 = new int [MAX_CHAR];
int [] count2 = new int [MAX_CHAR];
for ( int i = 0; i < n; i++)
{
count1[s1[i] - 'a' ] += 1;
count2[s2[i] - 'a' ] += 1;
}
int res = 0;
for ( int i = 0; i < MAX_CHAR; i++)
{
if (count1[i] != 0) {
res += Math.Abs(count1[i]
- count2[i]);
}
}
return res;
}
static public void Main()
{
string s1 = "abc" ;
string s2 = "cdd" ;
int len = s1.Length;
int res = leastCount(s1, s2, len);
Console.WriteLine(res);
}
}
|
Javascript
<script>
let MAX_CHAR = 26;
function leastCount(s1, s2, n)
{
let count1 = new Array(MAX_CHAR);
let count2 = new Array(MAX_CHAR);
for (let i = 0; i < MAX_CHAR; i++)
{
count1[i] = 0;
count2[i] = 0;
}
for (let i = 0; i < n; i++)
{
count1[s1[i].charCodeAt(0) -
'a' .charCodeAt(0)] += 1;
count2[s2[i].charCodeAt(0) -
'a' .charCodeAt(0)] += 1;
}
let res = 0;
for (let i = 0; i < MAX_CHAR; i++)
{
if (count1[i] != 0)
{
res += Math.abs(count1[i] - count2[i]);
}
}
return res;
}
let s1 = "abc" ;
let s2 = "cdd" ;
let len = s1.length;
let res = leastCount(s1, s2, len);
document.write(res);
</script>
|
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...