Consider n points on the Cartesian Coordinate Plane. Let point (Xi, Yi) is given a value Vi. A line parallel to y-axis is said to be a good partition line if sum of the point values on its left is equal to the sum of the point values on its right. Note that if a point lies on the partition line then it is not considered on either side of the line. The task is to print Yes if a good partition line exist else print No.
Example:
Input: X[] = {-3, 5, 8}, Y[] = {8, 7, 9}, V[] = {8, 2, 10}
Output: Yes
x = c where 5 < c < 8 is a good partition
line as sum of values on both sides is 10.Input: X[] = {-2, 5, 8, 9}, Y[] = {7, 7, 9, 8}, V[] = {6, 2, 1, 8}
Output: Yes
Approach:
- Count the value at each x-coordinate. This means that if there are multiple points with the same x-coordinate then they will be treated as a single point and their values will be added.
- Start from the minimum x-coordinate and check whether one of the following conditions is satisfied at Xi:
- Sum of the values till i – 1 is equal to sum of the values from i + 1 till n.
- Sum of the values till i is equal to sum of the values from i + 1 till n.
- Sum of the values till i – 1 is equal to sum of the values from i till n.
- If any of the above condition is satisfied for any given point then the line exists.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; const int MAX = 1000; // Function that returns true if // the required line exists bool lineExists(int x[], int y[], int v[], int n) { // To handle negative values from x[] int size = (2 * MAX) + 1; long arr[size] = { 0 }; // Update arr[] such that arr[i] contains // the sum of all v[j] such that x[j] = i // for all valid values of j for (int i = 0; i < n; i++) { arr[x[i] + MAX] += v[i]; } // Update arr[i] such that arr[i] contains // the sum of the subarray arr[0...i] // from the original array for (int i = 1; i < size; i++) arr[i] += arr[i - 1]; // If all the points add to 0 then // the line can be drawn anywhere if (arr[size - 1] == 0) return true; // If the line is drawn touching the // leftmost possible points if (arr[size - 1] - arr[0] == 0) return true; for (int i = 1; i < size - 1; i++) { // If the line is drawn just before // the current point if (arr[i - 1] == arr[size - 1] - arr[i - 1]) return true; // If the line is drawn touching // the current point if (arr[i - 1] == arr[size - 1] - arr[i]) return true; // If the line is drawn just after // the current point if (arr[i] == arr[size - 1] - arr[i]) return true; } // If the line is drawn touching the // rightmost possible points if (arr[size - 2] == 0) return true; return false; } // Driver code int main() { int x[] = { -3, 5, 8 }; int y[] = { 8, 7, 9 }; int v[] = { 8, 2, 10 }; int n = sizeof(x) / sizeof(int); if (lineExists(x, y, v, n)) cout << "Yes"; else cout << "No"; return 0; } |
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Java
// Java implementation of the approach class GFG { static int MAX = 1000; // Function that returns true if // the required line exists static boolean lineExists(int x[], int y[], int v[], int n) { // To handle negative values from x[] int size = (2 * MAX) + 1; long []arr = new long[size]; // Update arr[] such that arr[i] contains // the sum of all v[j] such that x[j] = i // for all valid values of j for (int i = 0; i < n; i++) { arr[x[i] + MAX] += v[i]; } // Update arr[i] such that arr[i] contains // the sum of the subarray arr[0...i] // from the original array for (int i = 1; i < size; i++) arr[i] += arr[i - 1]; // If all the points add to 0 then // the line can be drawn anywhere if (arr[size - 1] == 0) return true; // If the line is drawn touching the // leftmost possible points if (arr[size - 1] - arr[0] == 0) return true; for (int i = 1; i < size - 1; i++) { // If the line is drawn just before // the current point if (arr[i - 1] == arr[size - 1] - arr[i - 1]) return true; // If the line is drawn touching // the current point if (arr[i - 1] == arr[size - 1] - arr[i]) return true; // If the line is drawn just after // the current point if (arr[i] == arr[size - 1] - arr[i]) return true; } // If the line is drawn touching the // rightmost possible points if (arr[size - 2] == 0) return true; return false; } // Driver code public static void main(String []args) { int x[] = { -3, 5, 8 }; int y[] = { 8, 7, 9 }; int v[] = { 8, 2, 10 }; int n = x.length; if (lineExists(x, y, v, n)) System.out.printf("Yes"); else System.out.printf("No"); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach MAX = 1000; # Function that returns true if # the required line exists def lineExists(x, y, v, n) : # To handle negative values from x[] size = (2 * MAX) + 1; arr = [0] * size ; # Update arr[] such that arr[i] contains # the sum of all v[j] such that x[j] = i # for all valid values of j for i in range(n) : arr[x[i] + MAX] += v[i]; # Update arr[i] such that arr[i] contains # the sum of the subarray arr[0...i] # from the original array for i in range(1, size) : arr[i] += arr[i - 1]; # If all the points add to 0 then # the line can be drawn anywhere if (arr[size - 1] == 0) : return True; # If the line is drawn touching the # leftmost possible points if (arr[size - 1] - arr[0] == 0) : return True; for i in range(1, size - 1) : # If the line is drawn just before # the current point if (arr[i - 1] == arr[size - 1] - arr[i - 1]) : return True; # If the line is drawn touching # the current point if (arr[i - 1] == arr[size - 1] - arr[i]) : return True; # If the line is drawn just after # the current point if (arr[i] == arr[size - 1] - arr[i]) : return True; # If the line is drawn touching the # rightmost possible points if (arr[size - 2] == 0) : return True; return False; # Driver code if __name__ == "__main__" : x = [ -3, 5, 8 ]; y = [ 8, 7, 9 ]; v = [ 8, 2, 10 ]; n = len(x); if (lineExists(x, y, v, n)) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
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C#
// C# implementation of the approach using System; class GFG { static int MAX = 1000; // Function that returns true if // the required line exists static Boolean lineExists(int []x, int []y, int []v, int n) { // To handle negative values from x[] int size = (2 * MAX) + 1; long []arr = new long[size]; // Update arr[] such that arr[i] contains // the sum of all v[j] such that x[j] = i // for all valid values of j for (int i = 0; i < n; i++) { arr[x[i] + MAX] += v[i]; } // Update arr[i] such that arr[i] contains // the sum of the subarray arr[0...i] // from the original array for (int i = 1; i < size; i++) arr[i] += arr[i - 1]; // If all the points add to 0 then // the line can be drawn anywhere if (arr[size - 1] == 0) return true; // If the line is drawn touching the // leftmost possible points if (arr[size - 1] - arr[0] == 0) return true; for (int i = 1; i < size - 1; i++) { // If the line is drawn just before // the current point if (arr[i - 1] == arr[size - 1] - arr[i - 1]) return true; // If the line is drawn touching // the current point if (arr[i - 1] == arr[size - 1] - arr[i]) return true; // If the line is drawn just after // the current point if (arr[i] == arr[size - 1] - arr[i]) return true; } // If the line is drawn touching the // rightmost possible points if (arr[size - 2] == 0) return true; return false; } // Driver code public static void Main(String []args) { int []x = { -3, 5, 8 }; int []y = { 8, 7, 9 }; int []v = { 8, 2, 10 }; int n = x.Length; if (lineExists(x, y, v, n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by Rajput-Ji |
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Output:
Yes
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