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 pairslong 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 Codeint 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 approachimport 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 approachfrom collections import defaultdictimport math# Function to find the number of possible# ordered pairsdef 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 Codeif __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 approachusing 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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