Given an array arr[] of size N and an integer X, the task is to find the number of ordered pairs (i, j) such that i != j and X is the concatenation of the numbers arr[i] and arr[j]
Examples:
Input: N = 4, arr[] = {1, 212, 12, 12}, X = 1212
Output: 3
Explanation: X = 1212 can be obtained by concatenating:
- numbers[0] = 1 with numbers[1] = 212
- numbers[2] = 12 with numbers[3] = 12
- numbers[3] = 12 with numbers[2] = 12
- Therefore, number of possible ordered pairs are 3
Input: N = 3, arr[] = {11, 11, 110}, X = 11011
Output: 2
Approach: The task can be solved using a hashmap Follow the below steps to solve the problem
- Store the lengths of all the numbers in a vector.
- Iterate over the given number array and check if X can be obtained from the number. If so increase the count to check with how many numbers does the current number can form a pair.
- Increase the value of the current number in the map.
- Clear the map and repeat the same steps on the given number array from the back.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the number of possible // ordered pairs long long countPairs( int n, int x, vector< int > v) { int power[10] = { 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000 }; // Stores the count of digits of each number vector< int > len; long long count = 0; for ( int i = 0; i < n; i++) len.push_back( log10 (v[i]) + 1); unordered_map< int , int > mp; mp[v[0]]++; // Iterating from the start for ( int i = 1; i < n; i++) { if (x % power[len[i]] == v[i]) count += mp[x / power[len[i]]]; cout<< "count = " <<count<<endl; mp[v[i]]++; } mp.clear(); mp[v[n - 1]]++; // Iterating from the end for ( int i = n - 2; i >= 0; i--) { if (x % power[len[i]] == v[i]) count += mp[x / power[len[i]]]; cout<< "c = " <<count<<endl; mp[v[i]]++; } return count; } // Driver Code int main() { int N = 4; int X = 1212; vector< int > numbers = { 1, 212, 12, 12 }; long long ans = countPairs(N, X, numbers); cout << ans << endl; return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to find the number of possible // ordered pairs static long countPairs( int n, int x, int [] v) { int power[] = { 1 , 10 , 100 , 1000 , 10000 , 100000 , 1000000 , 10000000 , 100000000 , 1000000000 }; // Stores the count of digits of each number Vector<Integer> len = new Vector<Integer>(); long count = 0 ; for ( int i = 0 ; i < n; i++) len.add(( int ) (Math.log10(v[i]) + 1 )); HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); if (mp.containsKey(v[ 0 ])) { mp.put(v[ 0 ], mp.get(v[ 0 ]) + 1 ); } else { mp.put(v[ 0 ], 1 ); } // Iterating from the start for ( int i = 1 ; i < n; i++) { if (x % power[len.get(i)] == v[i]&&mp.containsKey(x / power[len.get(i)])) count += mp.get(x / power[len.get(i)]); System.out.println( "count = " + count); if (mp.containsKey(v[i])) { mp.put(v[i], mp.get(v[i]) + 1 ); } else { mp.put(v[i], 1 ); } } mp.clear(); if (mp.containsKey(v[n - 1 ])) { mp.put(v[n - 1 ], mp.get(v[n - 1 ]) + 1 ); } else { mp.put(v[n - 1 ], 1 ); } // Iterating from the end for ( int i = n - 2 ; i >= 0 ; i--) { if (x % power[len.get(i)] == v[i]&&mp.containsKey(x / power[len.get(i)])) count += mp.get(x / power[len.get(i)]); System.out.println( "c = " + count); if (mp.containsKey(v[i])) { mp.put(v[i], mp.get(v[i]) + 1 ); } else { mp.put(v[i], 1 ); } } return count; } // Driver Code public static void main(String[] args) { int N = 4 ; int X = 1212 ; int [] numbers = { 1 , 212 , 12 , 12 }; long ans = countPairs(N, X, numbers); System.out.print(ans + "\n" ); } } // This code is contributed by shikhasingrajput |
Python3
# Python 3 program for the above approach from collections import defaultdict import math # Function to find the number of possible # ordered pairs def countPairs(n, x, v): power = [ 1 , 10 , 100 , 1000 , 10000 , 100000 , 1000000 , 10000000 , 100000000 , 1000000000 ] # Stores the count of digits of each number length = [] count = 0 for i in range (n): length.append( int (math.log10(v[i])) + 1 ) mp = defaultdict( int ) mp[v[ 0 ]] + = 1 # Iterating from the start for i in range ( 1 , n): if (x % power[length[i]] = = v[i]): count + = mp[x / / power[length[i]]] mp[v[i]] + = 1 mp.clear() mp[v[n - 1 ]] + = 1 # Iterating from the end for i in range (n - 2 , - 1 , - 1 ): if (x % power[length[i]] = = v[i]): count + = mp[x / / power[length[i]]] mp[v[i]] + = 1 return count # Driver Code if __name__ = = "__main__" : N = 4 X = 1212 numbers = [ 1 , 212 , 12 , 12 ] ans = countPairs(N, X, numbers) print (ans) # This code is contributed by ukasp. |
C#
// C# program to implement above approach using System; using System.Collections; using System.Collections.Generic; class GFG { // Function to find the number of possible // ordered pairs static long countPairs( int n, int x, int [] v) { int [] power = new int []{ 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000 }; // Stores the count of digits of each number List< int > len = new List< int >(); long count = 0; for ( int i = 0 ; i < n ; i++){ len.Add(( int )(Math.Log10(v[i]) + 1)); } Dictionary< int , int > mp = new Dictionary< int , int >(); if (mp.ContainsKey(v[0])) { mp[v[0]]+=1; } else { mp.Add(v[0], 1); } // Iterating from the start for ( int i = 1; i < n; i++) { if (x % power[len[i]] == v[i] && mp.ContainsKey(x / power[len[i]])){ count += mp[x / power[len[i]]]; } // Console.WriteLine("count = " + count); if (mp.ContainsKey(v[i])) { mp[v[i]]+=1; } else { mp.Add(v[i], 1); } } mp.Clear(); if (mp.ContainsKey(v[n - 1])) { mp[v[n - 1]] += 1; } else { mp.Add(v[n - 1], 1); } // Iterating from the end for ( int i = n - 2; i >= 0; i--) { if (x % power[len[i]] == v[i] && mp.ContainsKey(x / power[len[i]])){ count += mp[x / power[len[i]]]; } // Console.WriteLine("c = " + count); if (mp.ContainsKey(v[i])) { mp[v[i]] += 1; } else { mp.Add(v[i], 1); } } return count; } public static void Main( string [] args){ int N = 4; int X = 1212; int [] numbers = new int []{ 1, 212, 12, 12 }; long ans = countPairs(N, X, numbers); Console.WriteLine(ans); } } // This code is contributed by subhamgoyal2014. |
Javascript
<script> // JavaScript code for the above approach // Function to find the number of possible // ordered pairs function countPairs(n, x, v) { let power = [1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000]; // Stores the count of digits of each number let len = []; let count = 0; for (let i = 0; i < n; i++) len.push(Math.floor(Math.log10(v[i])) + 1); let mp = new Map(); mp.set(v[0], 1); // Iterating from the start for (let i = 1; i < n; i++) { if (x % power[len[i]] == v[i] && mp.has(Math.floor(x / power[len[i]]))) count += mp.get(Math.floor(x / power[len[i]])); if (mp.has(v[i])) { mp.set(v[i], mp.get(v[i]) + 1) } else { mp.set(v[i], 1) } } mp = new Map(); mp.set(v[n - 1], 1); // Iterating from the end for (let i = n - 2; i >= 0; i--) { if (x % power[len[i]] == v[i] && mp.has(Math.floor(x / power[len[i]]))) count += mp.get(Math.floor(x / power[len[i]])); if (mp.has(v[i])) { mp.set(v[i], mp.get(v[i]) + 1) } else { mp.set(v[i], 1) } } return count; } // Driver Code let N = 4; let X = 1212; let numbers = [1, 212, 12, 12]; let ans = countPairs(N, X, numbers); document.write(ans + '<br>' ) // This code is contributed by Potta Lokesh </script> |
Output
3
Time complexity: O(N)
Auxiliary Space: O(N)
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