Sherlock and Anagrams


Problem Statement :


Two strings are anagrams of each other if the letters of one string can be rearranged to form the other string. Given a string, find the number of pairs of substrings of the string that are anagrams of each other.

For example s = mom , the list of all anagrammatic pairs is [m,m], [mo, om] at positions [ [1], [2], [0,1], [1,2] ] respectively .

Function Description

Complete the function sherlockAndAnagrams in the editor below. It must return an integer that represents the number of anagrammatic pairs of substrings in s.

sherlockAndAnagrams has the following parameter(s)
     s: a string .

Input Format

The first line contains an integer q , the number of queries .
Each of the next  q lines contains a string s to analyze.

Constraints
  
   1 <= q <= 10
   2 <= | s| <= 100


Output Format

For each query, return the number of unordered anagrammatic pairs.


Solution :



title-img 


                            Solution in C :

In C :


#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

typedef struct node{
    int s;
    int e;
}node;

void combine(node *A,int start,int end,int index,int *sum,node *temp,char *str){
    
    if(index == 2){
        int m[26] = {0};
        int i;
        for(i=temp[0].s;i<=temp[0].e;i++)
            m[str[i] - 'a']++;
        
        for(i=temp[1].s;i<=temp[1].e;i++)
            m[str[i] - 'a']--;
        
        for(i=0;i<26;i++)
            if(m[i])
                return;
            
        *sum += 1;
        return;
    }
    
    int i;
    
    for(i=start;i<end;i++){
        temp[index] = A[i];
        combine(A,i+1,end,index+1,sum,temp,str);
    }
    
}

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    int t;
    scanf("%d",&t);
    
    while(t--){
        char *str = (char*)malloc(sizeof(char)*101);
        scanf("%s",str);
        
        node *A = (node*)malloc(sizeof(node)*100);
        
        int index = 0;
        
        int i;
        int len = strlen(str);
        
        int count;
        int sum = 0;
        
        while(index<len){
            count = 0;
            for(i=0;i<len;i++){
                A[i].s = i;
                if(i+index>=len)
                    break;
                A[i].e = i+index;
                count++;
            }
            
            if(count>=2){
                node *temp = (node*)malloc(sizeof(node)*2);
                combine(A,0,count,0,&sum,temp,str);
            }
            index++;
        }
        
        printf("%d\n",sum);
        
    }
    return 0;
}
                        


                        Solution in C++ :

In C ++ :

#include<bits/stdc++.h>
#include <cstdio>
#define MAX 5000
using namespace std;
map<string,int> mp ;
int main(){
	ios::sync_with_stdio(0);
	int t;
	cin>>t;
	while(t--){
		mp.clear();
		string s,sn,ss ;
		int j;
		cin>>s;
		int l=s.length();
		for(int k=0;k<l;k++){
			ss = "";
			for(int i=0;i<l-k;i++){	
					j = k+i;
					ss = ss + s[j];
					sn = ss ;
					sort(sn.begin() , sn.end());
					mp[sn]++;
			}
		}
		long long int ans = 0 ;
		map<string,int> :: iterator it ;
		for(it = mp.begin() ; it != mp.end() ; it++){
			long long  vl = (long long)(it->second) ;
			if(vl > 1){				
				ans += (vl*(vl-1))/2LL ;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}
                    


                        Solution in Java :

In Java :


import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    
    private static int count[][] = new int[128][110];

    private static void resetCount() {
        for (int i = 0; i < count.length; i++) for (int j = 0; j < count[i].length; j++) count[i][j] = 0;
    }
    
    private static boolean areAnagrams(int from1, int to1, int from2, int to2) {
        for (int i = 'a'; i <= 'z'; i++) {
            if (count[i][to1+1]-count[i][from1] != count[i][to2+1]-count[i][from2])
                return false;
        }
        return true;
    }
    
    public static void main(String[] args) {
        final Scanner sc = new Scanner(System.in);
        final int TC = Integer.parseInt(sc.nextLine());
        for (int tc = 0; tc < TC; tc++) {
            final char s[] = sc.nextLine().toCharArray();
            resetCount();
            count[s[0]][1] = 1;
            for (int i = 1; i < s.length; i++) {
                for (int j = 'a'; j <= 'z'; j++) count[j][i+1] = count[j][i];
                count[s[i]][i+1]++;
            }
            int res = 0;
            for (int len = 1; len <= s.length-1; len++) {
                for (int from = 0; from <= s.length-len; from++) {
                    for (int to = from+1; to <= s.length-len; to++) {
                        if (areAnagrams(from, from+len-1, to, to+len-1)) res++;
                    }
                }
            }
            System.out.println(res);
        }
    }
}
                    


                        Solution in Python : 
                            
In Python3 :


for _ in range(int(input())):
    was = dict()
    s = input()

    n = len(s)
    for i in range(n):
        for j in range(i, n):
            cur = s[i:j + 1]
            cur = ''.join(sorted(cur))
            was[cur] = was.get(cur, 0) + 1

    ans = 0
    for x in was:
        v = was[x]
        ans += (v * (v - 1)) // 2

    print(ans)


View More Similar Problems

AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

View Solution →

Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

View Solution →

Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →

Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →

Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →