Count ways to divide C in two parts and add to A and B to make A strictly greater than B
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 â 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 â 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 â A + 1) / 2).
- The division should be ceiling division, thus we can rewrite as addA = max(0, (C + B â 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 â 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 B int 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 code int main() { int A = 3, B = 5, C = 5; cout << countWays(A, B, C); return 0; } |
Java
// Java implementation of the above approach import 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 B def 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 code A = 3 B = 5 C = 5 print (countWays(A, B, C)) # This code is contributed by shivanisingh |
C#
// C# implementation of the above approach using 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 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; 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 B function 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 code var 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)