Given three integers N, M and P. The task is to count the number of possible ordered triplets of form (x, y, z) where
1 ≤ x ≤ N, 1 ≤ y ≤ M and 1 ≤ z ≤ P
Since this count can be very large, return the answer modulo 109 + 7.
Examples:
Input: N = 3, M = 3, P = 3
Output: 6
Explanation: The possible triplets are:
(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)Input: N = 1, M = 2, P = 3
Output: 1
Explanation: Only one triplet is possible (1, 2, 3)
Approach: The solution is based on the following observation.
- Say after sorting the three numbers in ascending order they are A, B and C respectively.
- So there are A choices to choose first element.
- Now this choice is not available for choosing the second one. For second one there are total (B – 1) choices.
- Now these two choices are not available for the last. So there are only (C – 2) choices for third.
- Therefore the total number of possibilities are A * (B – 1) * (C – 2).
Follow the steps mentioned below to implement the above observation.
- Sort the three inputs in ascending order. Let the sorted order be (N1, N2, N3).
- Now apply the formula derived from the observation to get the final answer.
- Return the final answer modulo 109 + 7.
Below is the implementation of the above approach.
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;const unsigned int mod = 1000000007;// Function to count the // total number of possible tripletslong long int solve(int N, int M, int P){ int nums[] = { N, M, P }; sort(nums, nums + 3); long long int ans = ((nums[0] * (nums[1] - 1)) % mod * (nums[2] - 2) % mod)% mod; return ans;}// Driver codeint main(){ int N = 3, M = 3, P = 3; long long int ans = solve(N, M, P); cout << ans << endl; return 0;} |
Java
// Java code to implement the above approach// Importing Arrays class from the utility classimport java.util.Arrays;class GFG{ public static long mod = 1000000007; // Function to count the // total number of possible triplets static long solve(int N, int M, int P) { int nums[] = { N, M, P }; Arrays.sort(nums); long ans = ((nums[0] * (nums[1] - 1)) % mod * (nums[2] - 2) % mod) % mod; return ans; } // Driver method public static void main(String[] args) { int N = 3, M = 3, P = 3; long ans = solve(N, M, P); System.out.println(ans); }}// This code is contributed by rakeshsahni |
Python3
# Python3 program for the above approach# Function to count the total number of# possible tripletsdef solve(N, M, P): mod = 1000000007 nums = [ N, M, P ] nums.sort() ans = ((nums[0] * (nums[1] - 1)) % mod * (nums[2] - 2) % mod) % mod return ans# Driver Codeif __name__ == "__main__": N, M, P = 3, 3, 3 ans = solve(N, M, P) print(ans)# This code is contributed by Abhishek Thakur. |
C#
// C# code to implement the above approach// Importing Arrays class from the utility classusing System;class GFG{ static long mod = 1000000007; // Function to count the // total number of possible triplets static long solve(int N, int M, int P) { int []nums = { N, M, P }; Array.Sort(nums); long ans = ((nums[0] * (nums[1] - 1)) % mod * (nums[2] - 2) % mod) % mod; return ans; } // Driver method public static void Main() { int N = 3, M = 3, P = 3; long ans = solve(N, M, P); Console.Write((ans)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach let mod = 1000000007; // Function to count the // total number of possible triplets function solve(N, M, P) { let nums = [N, M, P]; nums.sort(function (a, b) { return a - b }) let ans = ((nums[0] * (nums[1] - 1)) % mod * (nums[2] - 2) % mod) % mod; return ans; } // Driver code let N = 3, M = 3, P = 3; let ans = solve(N, M, P); document.write(ans + '<br>')// This code is contributed by Potta Lokesh </script> |
Output
6
Time Complexity: O(1)
Auxiliary Space: O(1)
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