Given three integers A, B and C, the task is to count the number of ways to divide C into two parts and add to A and B such that A is strictly greater than B.
Examples:
Input: A = 5, B = 3, C = 4
Output: 3
The possible values of A and B after dividing C are:
A = 7, B = 5 where C is divided into 2 and 2.
A = 8, B = 4 where C is divided into 3 and 1.
A – 9, B = 3 where C is divided into 4 and 0.
Input: A = 3, B = 5, C = 5
Output: 2
Approach: On observing carefully, the following relation is formed for this problem.
- Let addA and addB be added to A and B respectively.
- Therefore, addA + addB = C and it should satisfy the inequality A + addA > B + addB.
- Now, since addB = C – addA and put it in the inequality:
A + addA > B + (C - addA) or, 2addA > C + B - A or, 2addA >= C + B - A + 1 or, addA >= (C + B - A + 1) / 2
- Since addA must be non negative, addA = max(0, (C + B – A + 1) / 2).
- The division should be ceiling division, thus we can rewrite as addA = max(0, (C + B – A + 2) / 2).
- Let this value be equal to minAddA. Since all integer values addA from [minAddA, C], satisfies the relation A + addA > B + addB, so the required number of ways is equal to max(0, C – minAddA + 1).
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to count the number of ways to divide// C into two parts and add to A and B such// that A is strictly greater than Bint countWays(int A, int B, int C){ // Minimum value added to A to satisfy // the given relation int minAddA = max(0, (C + B - A + 2) / 2); // Number of different values of A, i.e., // number of ways to divide C int count_ways = max(C - minAddA + 1, 0); return count_ways;}// Driver codeint main(){ int A = 3, B = 5, C = 5; cout << countWays(A, B, C); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{ // Function to count the number of ways to divide // C into two parts and add to A and B such // that A is strictly greater than B static int countWays(int A, int B, int C) { // Minimum value added to A to satisfy // the given relation int minAddA = Math.max(0, (C + B - A + 2) / 2); // Number of different values of A, i.e., // number of ways to divide C int count_ways = Math.max(C - minAddA + 1, 0); return count_ways; } // Driver code public static void main(String args[]) { int A = 3, B = 5, C = 5; System.out.println(countWays(A, B, C)); }}// This code is contributed by AbhiThakur |
Python3
# Python3 implementation of the above approach# Function to count the number of ways to divide# C into two parts and add to A and B such# that A is strictly greater than Bdef countWays(A, B, C): # Minimum value added to A to satisfy # the given relation minAddA = max(0, (C + B - A + 2) // 2) # Number of different values of A, i.e., # number of ways to divide C count_ways = max(C - minAddA + 1, 0) return count_ways# Driver codeA = 3B = 5C = 5print(countWays(A, B, C))# This code is contributed by shivanisingh |
C#
// C# implementation of the above approachusing System; class GFG{// Function to count the number of ways to divide// C into two parts and add to A and B such// that A is strictly greater than Bstatic int countWays(int A, int B, int C){ // Minimum value added to A to satisfy // the given relation int minAddA = Math.Max(0, (C + B - A + 2) / 2); // Number of different values of A, i.e., // number of ways to divide C int count_ways = Math.Max(C - minAddA + 1, 0); return count_ways;}// Driver Codepublic static void Main(String[] args){ int A = 3, B = 5, C = 5; Console.Write(countWays(A, B, C));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// Javascript implementation of the above approach// Function to count the number of ways to divide// C into two parts and add to A and B such// that A is strictly greater than Bfunction countWays(A, B, C){ // Minimum value added to A to satisfy // the given relation var minAddA = Math.max(0, parseInt((C + B - A + 2) / 2)); // Number of different values of A, i.e., // number of ways to divide C var count_ways = Math.max(C - minAddA + 1, 0); return count_ways;}// Driver codevar A = 3, B = 5, C = 5;document.write( countWays(A, B, C));// This code is contributed by rutvik_56.</script> |
Output:
2
Time Complexity: O(1)
Auxiliary Space: O(1)
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