Check whether second string can be formed from characters of first string
Last Updated :
29 Sep, 2022
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Given two strings str1 and str2, check if str2 can be formed from str1
Example :
Input : str1 = geekforgeeks, str2 = geeks
Output : Yes
Here, string2 can be formed from string1.Input : str1 = geekforgeeks, str2 = and
Output : No
Here string2 cannot be formed from string1.Input : str1 = geekforgeeks, str2 = geeeek
Output : Yes
Here string2 can be formed from string1
as string1 contains 'e' comes 4 times in
string2 which is present in string1.
The idea is to count frequencies of characters of str1 in a count array. Then traverse str2 and decrease frequency of characters of str2 in the count array. If frequency of a characters becomes negative at any point, return false.
Below is the implementation of above approach :
// CPP program to check whether second string
// can be formed from first string
#include <bits/stdc++.h>
using namespace std;
const int MAX = 256;
bool canMakeStr2(string str1, string str2)
{
// Create a count array and count frequencies
// characters in str1.
int count[MAX] = {0};
for (int i = 0; i < str1.length(); i++)
count[str1[i]]++;
// Now traverse through str2 to check
// if every character has enough counts
for (int i = 0; i < str2.length(); i++)
{
if (count[str2[i]] == 0)
return false;
count[str2[i]]--;
}
return true;
}
// Driver Code
int main()
{
string str1 = "geekforgeeks";
string str2 = "for";
if (canMakeStr2(str1, str2))
cout << "Yes";
else
cout << "No";
return 0;
}
// CPP program to check whether second string// can be formed from first stringusing namespace std;const int MAX = 256;bool canMakeStr2(string str1, string str2){ // Create a count array and count frequencies // characters in str1. int count[MAX] = {0}; for (int i = 0; i < str1.length(); i++) count[str1[i]]++; // Now traverse through str2 to check // if every character has enough counts for (int i = 0; i < str2.length(); i++) { if (count[str2[i]] == 0) return false; count[str2[i]]--; } return true;}// Driver Codeint main(){ string str1 = "geekforgeeks"; string str2 = "for"; if (canMakeStr2(str1, str2)) cout << "Yes"; else cout << "No"; return 0;}// Java program to check whether second string
// can be formed from first string
class GFG {
static int MAX = 256;
static boolean canMakeStr2(String str1, String str2)
{
// Create a count array and count frequencies
// characters in str1.
int[] count = new int[MAX];
char []str3 = str1.toCharArray();
for (int i = 0; i < str3.length; i++)
count[str3[i]]++;
// Now traverse through str2 to check
// if every character has enough counts
char []str4 = str2.toCharArray();
for (int i = 0; i < str4.length; i++) {
if (count[str4[i]] == 0)
return false;
count[str4[i]]--;
}
return true;
}
// Driver Code
static public void main(String []args)
{
String str1 = "geekforgeeks";
String str2 = "for";
if (canMakeStr2(str1, str2))
System.out.println("Yes");
else
System.out.println("No");
}
}
# Python program to check whether second string
# can be formed from first string
def canMakeStr2(s1, s2):
# Create a count array and count
# frequencies characters in s1
count = {s1[i] : 0 for i in range(len(s1))}
for i in range(len(s1)):
count[s1[i]] += 1
# Now traverse through str2 to check
# if every character has enough counts
for i in range(len(s2)):
if (count.get(s2[i]) == None or count[s2[i]] == 0):
return False
count[s2[i]] -= 1
return True
# Driver Code
s1 = "geekforgeeks"
s2 = "for"
if canMakeStr2(s1, s2):
print("Yes")
else:
print("No")
// C# program to check whether second string
// can be formed from first string
using System;
class GFG {
static int MAX = 256;
static bool canMakeStr2(string str1, string str2)
{
// Create a count array and count frequencies
// characters in str1.
int[] count = new int[MAX];
for (int i = 0; i < str1.Length; i++)
count[str1[i]]++;
// Now traverse through str2 to check
// if every character has enough counts
for (int i = 0; i < str2.Length; i++) {
if (count[str2[i]] == 0)
return false;
count[str2[i]]--;
}
return true;
}
// Driver Code
static public void Main()
{
string str1 = "geekforgeeks";
string str2 = "for";
if (canMakeStr2(str1, str2))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.
<?php
// PHP program to check whether
// second string can be formed
// from first string
$MAX = 256;
function canMakeStr2($str1, $str2)
{
// Create a count array and
// count frequencies characters
// in str1.
$count = (0);
for ($i = 0; $i < strlen($str1); $i++)
// Now traverse through str2
// to check if every character
// has enough counts
for ($i = 0; $i < strlen($str2); $i++)
{
if ($count[$str2[$i]] == 0)
return -1;
}
return true;
}
// Driver Code
$str1 = "geekforgeeks";
$str2 = "for";
if (canMakeStr2($str1, $str2))
echo "Yes";
else
echo "No";
// This code is contributed by ajit
?>
<script>
// Javascript program to check whether second string
// can be formed from first string
function canMakeStr2(str1, str2)
{
// Create a count array and count frequencies
// characters in str1.
let count = {};
// count.fill(0);
for (let i = 0; i < str1.length; i++) {
if (str1[i] in count)
count[str1[i]]++;
else
count[str1[i]] = 1
}
// Now traverse through str2 to check
// if every character has enough counts
for (let i = 0; i < str2.length; i++) {
if (count[str2[i]] == 0 || !(str2[i] in count))
return false;
count[str2[i]]--;
}
return true;
}
let str1 = "geekforgeeks";
let str2 = "gggeek";
if (canMakeStr2(str1, str2))
document.write("Yes");
else
document.write("No");
</script>
Output
Yes
Time Complexity: O(n+m), where n and m are the length of the given strings.
Auxiliary Space: O(256), which is constant.
