Skip to content
Related Articles

Related Articles

Maximize value of a pair from two given arrays based on given conditions
  • Last Updated : 09 Feb, 2021

Given two arrays A[] and B[] consisting of N integers and an integer K, the task is to find the maximum value of B[i] + B[j] + abs(A[i] – A[j]) by choosing any pair (i, j) such that abs(A[i] – A[j]) ≤ K.

Examples:

Input: A[] = {5, 6, 9, 10}, B[] = {3, 0, 10, -10}, K = 1
Output: 4
Explanation:
Only two pairs can be chosen, i.e. (0, 1) and (2, 3), because abs(A[0] – A[1]) ≤ K and abs(A[2] – A[3]) ≤ K.
The value of (0, 1) pair is B[0] + B[1] + abs(A[0] – A[1]) = 3 + 0 + 1 = 4.
The value of (2, 3) pair is B[2] + B[3] + abs(A[2] – A[3]) = 10 + (-10) + 1 = 1.
Hence, the maximum value from all possible pairs is 4.

Input: A[] = {1, 2, 3, 4}, B[] = {0, 8, 6, 9}, K = 2
Output: 19

Naive Approach: The simplest approach is to generate all possible pairs from the given array and count those pairs that satisfy the given conditions. After checking all the pairs print the count of all pairs.



Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use Segment Trees, Binary Search, and Sorting of the array according to the value of array A[]. Observe that, from the given equation it is clear that B[i] + B[j] + abs(A[i] – A[j]) equals to any of the below values: 

  • B[i] + B[j] + (A[i] – A[j])
  • B[i] + B[j] + (A[j] – A[i])

Consider 2nd equation:

B[i] + B[j] + A[j] – A[i] = B[i] – A[i] + (B[j] + A[j])

  • Here, it is observed that for each i in the array A[], finding the right most index of value smaller than of equal to A[i] + K. Let the rightmost index be right.
  • Now, calculate B[i] – A[i] + max value of B[j] + A[j] where i + 1 <= j <= right. This will give the maximum value of selected pair.
  • To calculate the maximum value each time in a range, segment trees can be used.

Follow the below steps to solve the problem:

  • Sort all values, B[i] and B[i] + A[i] according to the values of array A[].
  • Initialize maxValue as INT_MIN to store the final maximum answer and initialize a range maximum query segment tree storing all values, A[i] + B[i].
  • Traverse the array A[] and perform the following:
    • For each element find the index, say right, such that abs(A[i] – A[right]) <= K using Binary Search.
    • Then, find the maximum value, of (A[j] + B[j]) where i+1 <= j <= right.
    • Update maxValue as maxValue = max(maxValue, B[i] – A[i] + A[j] + B[j]).
  • After the above steps, print the value of maxValue as the result.

Below is the implementation of the above approach:

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
 
import java.util.*;
 
// Clas to store the values of a[i],
// b[i], a[i] + b[i]
class triplet implements Comparable<triplet> {
 
    int a, b, c;
 
    // Constructor
    triplet(int a, int b, int c)
    {
        this.a = a;
        this.b = b;
        this.c = c;
    }
 
    // Sort values according to array
    public int compareTo(triplet o)
    {
        return this.a - o.a;
    }
}
 
class GFG {
 
    // Stores the segment tree
    static int seg[];
 
    // Function to find the maximum
    // possible value according to the
    // given condition
    public static void maxValue(
        int a[], int b[], int n, int k)
    {
        // Initialize triplet array
        triplet arr[] = new triplet[n];
 
        // Store the values a[i], b[i],
        // a[i] + b[i]
        for (int i = 0; i < n; i++) {
 
            arr[i] = new triplet(a[i], b[i],
                                 a[i] + b[i]);
        }
 
        // Sort the array according
        // to array a[]
        Arrays.sort(arr);
 
        // Build segment tree
        seg = new int[4 * n];
        build(arr, 0, 0, n - 1);
 
        // Intialise the maxvalue of
        // the selected pairs
        int maxvalue = Integer.MIN_VALUE;
 
        // Traverse the array
        for (int i = 0; i < n; i++) {
 
            // For each value find the
            // floor(arr[i] + k)
            int right = search(arr,
                               arr[i].a + k);
 
            // Find the maximum value of
            // the select pairs i and max
            // from (i + 1, right)
            if (right != -1) {
 
                maxvalue = Math.max(
                    maxvalue, arr[i].b - arr[i].a
                                  + getMax(arr, 0, 0, n - 1,
                                           i + 1, right));
            }
        }
 
        // Print the maximum value
        System.out.println(maxvalue);
    }
 
    // Function to search floor of
    // the current value
    public static int search(
        triplet arr[], int val)
    {
        // Initialise low and high values
        int low = 0, high = arr.length - 1;
        int ans = -1;
 
        // Perform Binary Search
        while (low <= high) {
            int mid = low + (high - low) / 2;
 
            // If the current value is
            // <= val then store the
            // candidate answer and
            // find for rigtmost one
            if (arr[mid].a <= val) {
                ans = mid;
                low = mid + 1;
            }
            else
                high = mid - 1;
        }
        return ans;
    }
 
    // Function to build segment tree
    public static void build(
        triplet arr[], int index,
        int s, int e)
    {
        // Base Case
        if (s == e) {
            seg[index] = arr[s].c;
            return;
        }
        int mid = s + (e - s) / 2;
 
        // Build the left and right
        // segment trees
        build(arr, 2 * index + 1, s, mid);
        build(arr, 2 * index + 2, mid + 1, e);
 
        // Update current index
        seg[index] = Math.max(seg[2 * index + 1],
                              seg[2 * index + 2]);
    }
 
    // Function to get maximum value
    // in the range [qs, qe]
    public static int getMax(
        triplet arr[], int index, int s,
        int e, int qs, int qe)
    {
        // If segement is not in range
        if (qe < s || e < qs)
            return Integer.MIN_VALUE / 2;
 
        // If segement is completely
        // inside the query
        if (s >= qs && e <= qe)
            return seg[index];
 
        // Calcualate the maximum value
        // in left and right half
        int mid = s + (e - s) / 2;
        return Math.max(
            getMax(arr, 2 * index + 1,
                   s, mid, qs, qe),
            getMax(arr, 2 * index + 2,
                   mid + 1, e, qs, qe));
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int N = 4, K = 1;
        int A[] = { 5, 6, 9, 10 };
        int B[] = { 3, 0, 10, -10 };
 
        // Function call
        maxValue(A, B, N, K);
    }
}

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
using System.Collections.Generic;
 
// Clas to store the values of a[i],
// b[i], a[i] + b[i]
public class triplet : IComparable<triplet> {
 
    public int a, b, c;
 
    // Constructor
    public triplet(int a, int b, int c)
    {
        this.a = a;
        this.b = b;
        this.c = c;
    }
 
    // Sort values according to array
    public int CompareTo(triplet o)
    {
        return this.a - o.a;
    }
}
 
public class GFG {
 
    // Stores the segment tree
    static int []seg;
 
    // Function to find the maximum
    // possible value according to the
    // given condition
    public  void maxValue(
        int []a, int []b, int n, int k)
    {
        // Initialize triplet array
        triplet []arr = new triplet[n];
 
        // Store the values a[i], b[i],
        // a[i] + b[i]
        for (int i = 0; i < n; i++) {
 
            arr[i] = new triplet(a[i], b[i],
                                 a[i] + b[i]);
        }
 
        // Sort the array according
        // to array []a
        Array.Sort(arr);
 
        // Build segment tree
        seg = new int[4 * n];
        build(arr, 0, 0, n - 1);
 
        // Intialise the maxvalue of
        // the selected pairs
        int maxvalue = int.MinValue;
 
        // Traverse the array
        for (int i = 0; i < n; i++) {
  
            // For each value find the
            // floor(arr[i] + k)
            int right = search(arr,
                               arr[i].a + k);
 
            // Find the maximum value of
            // the select pairs i and max
            // from (i + 1, right)
            if (right != -1) {
 
                maxvalue = Math.Max(
                    maxvalue, arr[i].b - arr[i].a
                                  + getMax(arr, 0, 0, n - 1,
                                           i + 1, right));
            }
        }
 
        // Print the maximum value
        Console.WriteLine(maxvalue);
    }
 
    // Function to search floor of
    // the current value 
    public int search(
        triplet []arr, int val)
    {
        // Initialise low and high values
        int low = 0, high = arr.Length - 1;
        int ans = -1;
 
        // Perform Binary Search
        while (low <= high) {
            int mid = low + (high - low) / 2;
 
            // If the current value is
            // <= val then store the
            // candidate answer and
            // find for rigtmost one
            if (arr[mid].a <= val) {
                ans = mid;
                low = mid + 1;
            }
            else
                high = mid - 1;
        }
        return ans;
    }
 
    // Function to build segment tree
    public static void build(
        triplet []arr, int index,
        int s, int e)
    {
        // Base Case
        if (s == e) {
            seg[index] = arr[s].c;
            return;
        }
        int mid = s + (e - s) / 2;
 
        // Build the left and right
        // segment trees
        build(arr, 2 * index + 1, s, mid);
        build(arr, 2 * index + 2, mid + 1, e);
 
        // Update current index
        seg[index] = Math.Max(seg[2 * index + 1],
                              seg[2 * index + 2]);
    }
 
    // Function to get maximum value
    // in the range [qs, qe]
    public static int getMax(
        triplet []arr, int index, int s,
        int e, int qs, int qe)
    {
        // If segement is not in range
        if (qe < s || e < qs)
            return int.MinValue / 2;
 
        // If segement is completely
        // inside the query
        if (s >= qs && e <= qe)
            return seg[index];
 
        // Calcualate the maximum value
        // in left and right half
        int mid = s + (e - s) / 2;
        return Math.Max(
            getMax(arr, 2 * index + 1,
                   s, mid, qs, qe),
            getMax(arr, 2 * index + 2,
                   mid + 1, e, qs, qe));
    }
 
    // Driver Code
    public static void Main(String []args)
    {
        int N = 4, K = 1;
        int []A = { 5, 6, 9, 10 };
        int []B = { 3, 0, 10, -10 };
 
        // Function call
        new GFG().maxValue(A, B, N, K);
    }
}
 
// This code is contributed by 29AjayKumar

chevron_right


 
 

Output: 

4

 

 

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :