Given a 2D array a[][] where each row consists of N vector coordinates of the form (xi, yi, zi) of an applied body. If the body is in equilibrium, print “YES“. Otherwise, print “NO“.
Examples:
Input: a[][] = {{4, 1, 7}, {-2, 4, -1}, {1, -5, -3}}
Output: NOInput: a[][] = {{3, -1, 7}, {-5, 2, -4}, {2, -1, -3}}
Output: YES
Approach: Follow the steps below to solve the problem:
- Apply vector addition on all the vector coordinates.
- If the result of the vector addition is (0, 0, 0), the body is in equilibrium. Therefore, print “YES“.
- Otherwise, print “NO“.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <iostream>using namespace std;// Function to calculate vector additionvoid vectorAddition(int a[][3], int N){ // Sum1 to store sum of xi // Sum2 to store sum of yi // Sum3 to store sum of zi int sum1 = 0, sum2 = 0, sum3 = 0; for (int i = 0; i < N; i++) { sum1 += a[i][0]; sum2 += a[i][1]; sum3 += a[i][2]; } // If the sum1, sum2 and sum3 // are all equal to zero if (sum1 == 0 && sum2 == 0 && sum3 == 0) { // Body is in // equilibrium cout << "YES"; } else { // Body is not in // equilibrium cout << "NO"; }}// Driver Codeint main(){ int N = 3; int a[N][3] = { { 3, -1, 7 }, { -5, 2, -4 }, { 2, -1, -3 } }; vectorAddition(a, N); return 0;} |
Java
// Java Program to implement// the above approachimport java.util.*;class GFG{// Function to calculate // vector additionstatic void vectorAddition(int a[][], int N){ // Sum1 to store sum of xi // Sum2 to store sum of yi // Sum3 to store sum of zi int sum1 = 0, sum2 = 0, sum3 = 0; for (int i = 0; i < N; i++) { sum1 += a[i][0]; sum2 += a[i][1]; sum3 += a[i][2]; } // If the sum1, sum2 and sum3 // are all equal to zero if (sum1 == 0 && sum2 == 0 && sum3 == 0) { // Body is in // equilibrium System.out.print("YES"); } else { // Body is not in // equilibrium System.out.print("NO"); }}// Driver Codepublic static void main(String[] args){ int N = 3; int a[][] = {{3, -1, 7}, {-5, 2, -4}, {2, -1, -3}}; vectorAddition(a, N);}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program to implement# the above approach# Function to calculate vector additiondef vectorAddition(a, N): # Sum1 to store sum of xi # Sum2 to store sum of yi # Sum3 to store sum of zi sum1 = 0 sum2 = 0 sum3 = 0 for i in range(N): sum1 += a[i][0] sum2 += a[i][1] sum3 += a[i][2] # If the sum1, sum2 and sum3 # are all equal to zero if (sum1 == 0 and sum2 == 0 and sum3 == 0): # Body is in # equilibrium print("YES") else: # Body is not in # equilibrium print("NO")# Driver Codeif __name__ == '__main__': N = 3 a = [ [ 3, -1, 7 ], [ -5, 2, -4 ], [ 2, -1, -3 ] ] vectorAddition(a, N)# This code is contributed by mohit kumar 29 |
C#
// C# Program to implement// the above approachusing System;class GFG{// Function to calculate // vector additionstatic void vectorAddition(int[,]a, int N){ // Sum1 to store sum of xi // Sum2 to store sum of yi // Sum3 to store sum of zi int sum1 = 0, sum2 = 0, sum3 = 0; for (int i = 0; i < N; i++) { sum1 += a[i, 0]; sum2 += a[i, 1]; sum3 += a[i, 2]; } // If the sum1, sum2 and sum3 // are all equal to zero if (sum1 == 0 && sum2 == 0 && sum3 == 0) { // Body is in // equilibrium Console.Write("YES"); } else { // Body is not in // equilibrium Console.Write("NO"); }}// Driver Codepublic static void Main(String[] args){ int N = 3; int[,]a = {{3, -1, 7}, {-5, 2, -4}, {2, -1, -3}}; vectorAddition(a, N);}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript program for the above approach// Function to calculate// vector additionfunction vectorAddition(a, N){ // Sum1 to store sum of xi // Sum2 to store sum of yi // Sum3 to store sum of zi let sum1 = 0, sum2 = 0, sum3 = 0; for (let i = 0; i < N; i++) { sum1 += a[i][0]; sum2 += a[i][1]; sum3 += a[i][2]; } // If the sum1, sum2 and sum3 // are all equal to zero if (sum1 == 0 && sum2 == 0 && sum3 == 0) { // Body is in // equilibrium document.write("YES"); } else { // Body is not in // equilibrium document.write("NO"); }}// Driver Code let N = 3; let a = [[3, -1, 7], [-5, 2, -4], [2, -1, -3]]; vectorAddition(a, N); </script></script> |
Output
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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