# Sparse Arrays

### Problem Statement :

```There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results.

Example:
strings=['ab', 'ab', 'abc']
queries=['ab', 'abc', 'bc']

There are  instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0].

Function Description:

Complete the function matchingStrings in the editor below. The function must return an array of integers representing the frequency of occurrence of each query string in strings.

matchingStrings has the following parameters:

1. string strings[n] - an array of strings to search
2. string queries[q] - an array of query strings

Returns:
1. int[q]: an array of results for each query

Input Format:

The first line contains and integer n, the size of strings[].
Each of the next n lines contains a string strings[i].
The next line contains q, the size of queries[].
Each of the next q lines contains a string queries[i].

Constraints:
1. 1<=n<=1000
2. 1<=q<=1000
3. 1<=strings[i],queries[i]<=20.```

### Solution :

```                            ```Solution in C :

In C:

#include<stdio.h>
#include<string.h>
int main()
{
char A,B;
int N,i,Q,ans;
scanf("%d",&N);
for(i=0;i<N;i++)
scanf("%s",A[i]);
scanf("%d",&Q);
while(Q--)
{
ans=0;
scanf("%s",B);
for(i=0;i<N;i++)
{
if(strcmp(B,A[i])==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}

In C++:

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<map>
#include<string>
using namespace std;

int main()
{

int n;
cin >> n;
map<string, int>mp;
while (n--)
{
string t;
cin >> t;
mp[t]++;
}
int a;
cin >> a;
while (a--)
{
string t;
cin >> t;
cout << mp[t] << endl;
}
return 0;
}

In Java:

import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) throws IOException {
HashMap<String,Integer> map = new HashMap<String,Integer>();
while(numstr > 0)
{
if (map.get(str)==null)
map.put(str, 1);
else
{
int num = map.get(str);
map.put(str, num+1);
}
numstr--;
}
while(numops > 0)
{
if (map.get(s) == null)
System.out.println(0);
else
System.out.println(map.get(s));
numops--;
}
}
}

In Python 3:

N = int(input())
numbers = list()
counts = list()
for i in range(0,N):
numbers.append(input())
Q = int(input())
for i in range(0,Q):
checkstring = input()
count = 0
for num in numbers:
if num == checkstring:
count = count + 1
counts.append(count)
for count in counts:
print(count)```
```

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

## Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty