Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find the Number of solutions for the equation x + y + z <= n using C++
In this article, we will explain the approaches to find the number of solutions for the equation x+y+z<=n. In this problem, we have an equation with four variables, and the task is to find the solution for the given equation. So here is a simple example &miuns;
Input: X = 1, Y = 1, Z = 1, n = 1 Output: 4 Input: X = 1, Y = 2, Z = 3, n = 4 Output: 3
In this problem, we can simply go through all the values of (x, y), (y,z), (x,z) by isolating each variable and checking if it satisfies the equation or not.
Approach to Find the Solution
Now we will use the Brute Force approach to find the solution to the given problem.
Brute Force
In this program we are going to go through all the possible values of (x,y), (y,z) and (x,z) such that it satisfies the equation z <= n - x - y (here z is isolated) where 0 <= z <= Z (and same for other isolated variables).
Example
#include<bits/stdc++.h>
using namespace std;
int main(){
int X = 1, Y = 2, Z = 3, n = 4; // limits of x, y, z and given n.
int answer = 0; // counter variable.
for(int i = 0; i <= X; i++){
for(int j = 0; j <= Y; j++){
int temp = (n - i) - j; // temp = n - x - y.
if(temp >= Z){ // if n - x - y >= z so we increment the answer.
answer++;
}
}
}
for(int i = 0; i <= X; i++){
for(int j = 0; j <= Z; j++){
int temp = (n - i) - j; // temp = n - x - y.
if(temp >= Y){ // if n - x - y >= z so we increment the answer.
answer++;
}
}
}
for(int i = 0; i <= Z; i++){
for(int j = 0; j <= Y; j++){
int temp = (n - i) - j; // temp = n - x - y.
if(temp >= X){ // if n - x - y >= z so we increment the answer.
answer++;
}
}
}
cout << answer << "\n";
}Output
17
Explanation of the Above Program
In this program, we are going to go through all the combinations of (x,y), (y, z), (x,z) using a nested for loop and check the equation if it satisfies the equation or not and if it satisfies then, we increment the answer.
Conclusion
In this article, we solve a problem finding the Number of solutions that satisfy the equation x + y + z<= n in O(X*Y) time complexity. We also learned the C++ program for this problem and the complete approach by which we solved this problem. We can write the same program in other languages such as C, java, python, and other languages.
199 Views
