Maximum the value of a given expression for any pair of coordinates on a 2D plane
Last Updated :
10 Jan, 2023
Given a sorted 2D array arr[][2] of size N such that (arr[i][0], arr[i][1]) represents the coordinates of ith point in the cartesian plane and an integer K, the task is to find the maximum value of the expression (|arr[i][0] - arr[j][0]| + arr[i][1] + arr[j][1]) such that |arr[i][0] - arr[j][0]| ? K for any possible pair of coordinates (i, j).
Examples:
Input: arr[][] = {{1, 3}, {2, 0}, {5, 10}, {6, -10}}, K = 1
Output: 4
Explanation:
Choose pairs (0, 1). Now the value of the expression is given by:
value = (abs(1 - 2) + 3 + 0) = 4, which is maximum and abs(1 - 2) = 1(? K).
Therefore, print 4.
Input: arr[][] = {{0, 0}, {3, 0}, {9, 2}}, K = 3
Output: 3
Approach: The given problem can be solved using a Greedy Algorithm using the priority queue which is based on the following observations:
- Rearranging the expression for all i > j as (arr[i][0] - arr[i][1] + arr[j][0] + arr[j][1]).
- Now, keeping the pair of {arr[i]x - arr[i]y, arr[i]x} in sorted order, the value of the given expression for every array element at index j can be calculated.
Follow the steps below to solve the problem:
- Initialize a priority_queue of pairs say PQ that stores the pair of differences of coordinated axes of a point and X coordinate of that point.
- Initialize a variable say res as INT_MIN to store the maximum value.
- Traverse the array arr[][] and considering {X, Y} is the current point perform the following operations:
- Iterate while PQ is not empty and (X - PQ.top()[1]) is greater than K and remove the top element from the priority_queue PQ.
- If PQ is not empty then update the value of res as the maximum of res and PQ.top()[0] + X + Y).
- Push the pair {Y - X, X} into the priority_queue PQ.
- After completing the above steps, print the value of res as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum value
// of the given expression possible
// for any pair of co-ordinates
void findMaxValueOfEquation(
vector<vector<int> >& arr, int K)
{
// Stores the differences between pairs
priority_queue<vector<int> > pq;
// Stores the maximum value
int res = INT_MIN;
// Traverse the array arr[][]
for (auto point : arr) {
// While pq is not empty and
// difference between point[0]
// and pq.top()[1] > K
while (!pq.empty()
&& point[0] - pq.top()[1]
> K) {
// Removes the top element
pq.pop();
}
// If pq is not empty
if (!pq.empty()) {
// Update the value res
res = max(res,
pq.top()[0] + point[0] + point[1]);
}
// Push pair {point[1] - point[0],
// point[0]} in pq
pq.push({ point[1] - point[0],
point[0] });
}
// Print the result
cout << res;
}
// Driver Code
int main()
{
vector<vector<int> > arr
= { { 1, 3 }, { 2, 0 },
{ 5, 10 }, { 6, -10 } };
int K = 1;
findMaxValueOfEquation(arr, K);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG
{
// Function to find the maximum value
// of the given expression possible
// for any pair of co-ordinates
static void findMaxValueOfEquation(int arr[][], int K)
{
// Stores the differences between pairs
PriorityQueue<int[]> pq
= new PriorityQueue<>((a, b) -> {
if (a[0] != b[0])
return b[0] - a[0];
return b[1] - a[1];
});
// Stores the maximum value
int res = Integer.MIN_VALUE;
// Traverse the array arr[][]
for (int point[] : arr) {
// While pq is not empty and
// difference between point[0]
// and pq.top()[1] > K
while (!pq.isEmpty()
&& point[0] - pq.peek()[1] > K) {
// Removes the top element
pq.poll();
}
// If pq is not empty
if (!pq.isEmpty()) {
// Update the value res
res = Math.max(res, pq.peek()[0] + point[0]
+ point[1]);
}
// Push pair {point[1] - point[0],
// point[0]} in pq
pq.add(new int[] { point[1] - point[0],
point[0] });
}
// Print the result
System.out.println(res);
}
// Driver Code
public static void main(String[] args)
{
int[][] arr
= { { 1, 3 }, { 2, 0 }, { 5, 10 }, { 6, -10 } };
int K = 1;
findMaxValueOfEquation(arr, K);
}
}
// This code is contributed by Kingash.
# Python3 program for the above approach
# Function to find the maximum value
# of the given expression possible
# for any pair of co-ordinates
def findMaxValueOfEquation(arr, K):
# Stores the differences between pairs
pq = []
# Stores the maximum value
res = -10**8
# Traverse the array arr[][]
for point in arr:
# While pq is not empty and
# difference between point[0]
# and pq.top()[1] > K
while (len(pq)>0 and point[0] - pq[-1][1] > K):
# Removes the top element
del pq[-1]
# If pq is not empty
if (len(pq) > 0):
# Update the value res
res = max(res, pq[-1][0] + point[0] + point[1])
# Push pair {point[1] - point[0],
# point[0]} in pq
pq.append([point[1] - point[0], point[0]])
pq = sorted(pq)
# Print the result
print (res)
# Driver Code
if __name__ == '__main__':
arr = [ [ 1, 3 ], [ 2, 0 ], [ 5, 10 ], [ 6, -10 ] ]
K = 1
findMaxValueOfEquation(arr, K)
# This code is contributed by mohit kumar 29.
// C# code to implement the approach
using System;
using System.Collections.Generic;
class Program {
static void FindMaxValueOfEquation(int[][] arr, int K)
{
// Stores the differences between pairs
var pq = new SortedSet<int[]>(
Comparer<int[]>.Create((a, b) => {
if (a[0] != b[0]) {
return b[0] - a[0];
}
return b[1] - a[1];
}));
// Stores the maximum value
int res = int.MinValue;
// Traverse the array arr[][]
foreach(int[] point in arr)
{
// While pq is not empty and
// difference between point[0]
// and pq.top()[1] > K
while (pq.Count > 0
&& point[0] - pq.Min[1] > K) {
// Removes the top element
pq.Remove(pq.Min);
}
// If pq is not empty
if (pq.Count > 0) {
// Update the value res
res = Math.Max(res, pq.Min[0] + point[0]
+ point[1]);
}
// Push pair {point[1] - point[0],
// point[0]} in pq
pq.Add(new int[] { point[1] - point[0],
point[0] });
}
// Print the result
Console.WriteLine(res);
}
static void Main(string[] args)
{
int[][] arr
= { new int[] { 1, 3 }, new int[] { 2, 0 },
new int[] { 5, 10 }, new int[] { 6, -10 } };
int K = 1;
FindMaxValueOfEquation(arr, K);
}
}
// This code is contributed by phasing17
<script>
// Javascript program for the above approach
// Function to find the maximum value
// of the given expression possible
// for any pair of co-ordinates
function findMaxValueOfEquation(arr,K)
{
// Stores the differences between pairs
let pq=[];
// Stores the maximum value
let res = Number.MIN_VALUE;
// Traverse the array arr[][]
for (let point=0;point< arr.length;point++) {
// While pq is not empty and
// difference between point[0]
// and pq.top()[1] > K
while (pq.length!=0
&& arr[point][0] - pq[0][1] > K) {
// Removes the top element
pq.shift();
}
// If pq is not empty
if (pq.length!=0) {
// Update the value res
res = Math.max(res, pq[0][0] + arr[point][0]
+ arr[point][1]);
}
// Push pair {point[1] - point[0],
// point[0]} in pq
pq.push([ arr[point][1] - arr[point][0],
arr[point][0] ]);
pq.sort(function(a,b){
if (a[0] != b[0])
return b[0] - a[0];
return b[1] - a[1];})
}
// Print the result
document.write(res);
}
// Driver Code
let arr = [[ 1, 3 ], [ 2, 0 ], [ 5, 10 ], [ 6, -10 ]];
let K = 1;
findMaxValueOfEquation(arr, K);
// This code is contributed by unknown2108
</script>
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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