CareerCup

#include<iostream>

bool checkIsPermutation(const char* a, const char* b)
{
	if (strlen(a) != strlen(b))
		return false;

	char arr[256];
	memset(arr, 0, sizeof(arr));

	for (int i = 0; i < strlen(a); i++)
	{
		arr[a[i]]++;
		arr[b[i]]--;
	}

	for (int i = 0; i < 256; i++)
	{
		if (arr[i] != 0)
			return false;
	}

	return true;
};

int main()
{
	char* a = "microsoft";
	char* b = "softmciro";

	bool res = checkIsPermutation(a, b);

	return 0;

}

1.take an array of 256 size,(for ascii charctrs),initialise with 0.
2.iterate through the string,increment the value of array by 1,on position pointd by ascii value of char obtained.
3.
do this for both string.
4.iterate through array of both strings,if same. than true else false..
time complexity o(n)
space complexity o(1){since int array is independent of string size}.

one solution is to sort those strings and compare.

bool isPermutation(string s1, string s2)
{
if(s1.length != s2.length)
return false;

int []arr = new int[26];

for(int i=0;i<s1.length;i++)
{
int index = s1[i] - 'a';
arr[index]++;

index = s2[i] - 'a';
arr[index]++;

}
for(int i=0;i<arr.length;i++)
if(arr[i] != 0)
return false;

returnt true;
}

Use built-in HashMap as <character, count> and check for both the strings.

This could also be done by traversing both the list and adding the values of each character,if the final answer is same then the strings are permutations of one another otherwise not.

#include<iostream>
#include<string.h>

using namespace std;

int main()
{
	char a[]="hello", b[]="oheee";
	if(strlen(a)!=strlen(b))
	{
		cout<<"\nNO\n";		
		return 0;	
	}				
	int count[53];
	for(int i=0;i<52;++i)
	count[i]=0;
	
	int k,i;
	k=strlen(a);
	i=0;
	while(i<k)
	{	
		count[(int)a[i]-65]+=1;		
		count[(int)b[i]-65]-=1;		
		i++;
	}
	
	i=0;	
	while(i<52)
	{
		if(count[i]!=0)
		{
			cout<<"\nNO\n";
			return 0;
		}	
		i++;
	}
	cout<<"\n YeS\n";
	return 0;
}

#include<string.h>
#include<conio.h>
#include<malloc.h>
#include<stdio.h>
#include<iostream.h>
#include<iostream>
void check(char *,char *);
void check(char *a,char *b)
{
if(strlen(a)!=strlen(b))
{
printf("It is not permutation");
}
else
{
int x=0,y=0;
for(int i=0;i<strlen(a);i++)
{
x=x+*(a+i);
y=y+*(b+i);
}
if(x==y)
{
printf("It is permutation");
}
else
{
printf("It is not a permutation");
}
}
}
int main()
{
char* a=(char *)malloc(100) ;
char* b=(char *)malloc(100) ;
gets(a);
gets(b);


check(a,b);

getch();

return 0;

}

Method1: traverse two strings, one from its head and the other from the end. compare every elements from both string.
Method2: first get two strings length( which will cost o(n) time); compare the lengths if the are not equal, return false; else traverse first string by increasing variable i, and compare str1[i] with str2[len-i-1] if one of them are not equal return false.
...

con cat both string original and permutation and check whether original one is sub string of concated one

#include<stdio.h>
#include<conio.h>
#include<string.h>
#define MAX 50
int main()
{
	char str1[MAX],str2[MAX];
	int i,j,count=0;
	scanf("%s%s",str1,str2);
	if(strlen(str1)!=strlen(str2))
	printf("%s and %s are not permutations of each other",str1,str2);
	else
	{
		for(i=0;i<strlen(str1);i++)
		{
			for(j=0;j<strlen(str2);j++)
			{
				if(str1[i]==str2[j])
				{
				count++;
				break;
			}}
		}
		if(count==strlen(str1))
		printf("%s and %s are permutations of each other",str1,str2);
		else
		printf("%s and %s are not permutations of each other",str1,str2);
	}
	return 0;
}

#include<stdio.h>
#include<string.h>
int main()
{
        char str1[10000];
        char str2[10000];
        scanf("%s",str1);
        scanf("%s",str2);
        int i,k1,k2,sum1=0,sum2=0;
        k1=strlen(str1);
        k2=strlen(str2);
        if(k1!=k2)
                printf("FALSE");
        else{
                for(i=0;i<k1;i++)
                {
                        sum1+=str1[i];
                        sum2+=str2[i];
                }
                if(sum1==sum2)
                        printf("TRUE");
                else
                        printf("FALSE");
        }
        return 0;
}