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:
#include <bits/stdc++.h> using namespace std;
// c++ code for the above approach // Class to store the values of a[i], b[i], a[i] + b[i] class Triplet {
public :
int a, b, c;
Triplet( int d, int e, int f) {
a = d;
b = e;
c = f;
}
}; // Stores the segment tree vector< int > seg;
// Function to search floor of // the current value int search(vector<Triplet> arr, int val)
{ // Initialise low and high values
int low = 0;
int high = arr.size() - 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 rightmost one
if (arr[mid].a <= val) {
ans = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
return ans;
} // Function to build segment tree void build(vector<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;
// Buildthe 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] = max(seg[2 * index + 1], seg[2 * index + 2]);
} // Function to get maximum value // in the range [qs, qe] int getMax(vector<Triplet> arr, int index, int s, int e, int qs, int qe)
{ // If segment is not in range
if (qe < s || e < qs)
{
return INT_MIN;
}
// If segment is fully in range
if (qs <= s && e <= qe) {
return seg[index];
}
// If a part of this segment is
// in range
int mid = s + (e - s) / 2;
int left = getMax(arr, 2 * index + 1, s, mid, qs, qe);
int right = getMax(arr, 2 * index + 2, mid + 1, e, qs, qe);
// Return the maximum value
return max(left, right);
} // Function to find the maximum // possible value according to the // given condition void maxValue(vector< int > a,vector< int > b, int n, int k)
{ // Initialize triplet array
vector<Triplet> arr;
// Store the values a[i], b[i],
// a[i] + b[i]
for ( int i = 0; i < n; i++) {
arr.push_back(Triplet(a[i], b[i], a[i] + b[i]));
}
// Sort the array according
// to array a[]
sort(arr.begin(), arr.end(), [] ( const Triplet &x, const Triplet &y){
return x.a - y.a;
});
// Build segment tree
seg.resize(4*n);
int x = 5;
build(arr, 0, 0, n - 1);
// Initialise the maxvalue of
// the selected pairs
int maxvalue = INT_MIN;
// 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 = max(maxvalue, arr[i].b - arr[i].a + getMax(arr, 0, 0, n - 1, i + 1, right));
}
}
// Print the maximum value
cout << maxvalue - x;
} // Driver code int main(){
vector< int > a = {5, 6, 9, 10};
vector< int > b = {3, 0, 10, -10};
int N = 4;
int K = 1;
maxValue(a, b, N, K);
return 0;
} // This code is contributed by Arushi Jindal. |
// Java program for the above approach import java.util.*;
// Class 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 );
// Initialise 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 rightmost 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 segment is not in range
if (qe < s || e < qs)
return Integer.MIN_VALUE / 2 ;
// If segment is completely
// inside the query
if (s >= qs && e <= qe)
return seg[index];
// Calculate 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);
}
} |
# Class to store the values of a[i], b[i], a[i] + b[i] class Triplet:
def __init__( self , a, b, c):
self .a = a
self .b = b
self .c = c
# Sort values according to array
def __lt__( self , other):
return self .a < other.a
# Stores the segment tree seg = []
# Function to find the maximum # possible value according to the # given condition def maxValue(a, b, n, k):
global seg
# Initialize triparray
arr = [Triplet( 0 , 0 , 0 ) for i in range (n)]
# Store the values a[i], b[i],
# a[i] + b[i]
for i in range (n):
arr[i] = Triplet(a[i], b[i], a[i] + b[i])
# Sort the array according
# to array a[]
arr.sort()
# Build segment tree
seg = [ 0 ] * ( 4 * n)
build(arr, 0 , 0 , n - 1 )
# Initialise the maxvalue of
# the selected pairs
maxvalue = float ( '-inf' )
# Traverse the array
for i in range (n):
# For each value find the
# floor(arr[i] + k)
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 = max (
maxvalue, arr[i].b - arr[i].a + getMax(arr, 0 , 0 , n - 1 , i + 1 , right))
# Print the maximum value
print (maxvalue)
# Function to search floor of # the current value def search(arr, val):
# Initialise low and high values
low = 0
high = len (arr) - 1
ans = - 1
# Perform Binary Search
while low < = high:
mid = low + (high - low) / / 2
# If the current value is
# <= val then store the
# candidate answer and
# find for rightmost one
if arr[mid].a < = val:
ans = mid
low = mid + 1
else :
high = mid - 1
return ans
# Function to build segment tree def build(arr, index, s, e):
# Base Case
if s = = e:
seg[index] = arr[s].c
return
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] = max (seg[ 2 * index + 1 ], seg[ 2 * index + 2 ])
# Function to get maximum value # in the range [qs, qe] def getMax(arr, index, s, e, qs, qe):
# If segment is not in range
if qe < s or e < qs:
return float ( '-inf' )
# If segment is fully in range
if qs < = s and e < = qe:
return seg[index]
# If a part of this segment is
# in range
mid = s + (e - s) / / 2
left = getMax(arr, 2 * index + 1 , s, mid, qs, qe)
right = getMax(arr, 2 * index + 2 , mid + 1 , e, qs, qe)
# Return the maximum value
return max (left, right)
# Driver code a = [ 5 , 6 , 9 , 10 ]
b = [ 3 , 0 , 10 , - 10 ]
N = 4
K = 1
maxValue(a, b, N, K) # This code is contributed by phasing17 |
// C# program for the above approach using System;
using System.Collections.Generic;
// Class 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);
// Initialise 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 rightmost 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 segment is not in range
if (qe < s || e < qs)
return int .MinValue / 2;
// If segment is completely
// inside the query
if (s >= qs && e <= qe)
return seg[index];
// Calculate 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 |
// JavaScript code for the above approach
// Class to store the values of a[i], b[i], a[i] + b[i]
class Triplet {
constructor(a, b, c) {
this .a = a;
this .b = b;
this .c = c;
}
// Sort values according to array
compareTo(other) {
return this .a - other.a;
}
}
// Stores the segment tree
let seg = [];
// Function to find the maximum
// possible value according to the
// given condition
function maxValue(a, b, n, k)
{
// Initialize triplet array
let arr = new Array(n);
// Store the values a[i], b[i],
// a[i] + b[i]
for (let i = 0; i < n; i++) {
arr[i] = new Triplet(a[i], b[i], a[i] + b[i]);
}
// Sort the array according
// to array a[]
arr.sort((a, b) => a.compareTo(b));
// Build segment tree
seg = new Array(4 * n);
build(arr, 0, 0, n - 1);
// Initialise the maxvalue of
// the selected pairs
let maxvalue = Number.MIN_SAFE_INTEGER;
// Traverse the array
for (let i = 0; i < n; i++)
{
// For each value find the
// floor(arr[i] + k)
let 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.log(maxvalue);
}
// Function to search floor of
// the current value
function search(arr, val)
{
// Initialise low and high values
let low = 0;
let high = arr.length - 1;
let ans = -1;
// Perform Binary Search
while (low <= high) {
let mid = low + Math.floor((high - low) / 2);
// If the current value is
// <= val then store the
// candidate answer and
// find for rightmost one
if (arr[mid].a <= val) {
ans = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
return ans;
}
// Function to build segment tree
function build(arr, index, s, e)
{
// Base Case
if (s === e) {
seg[index] = arr[s].c;
return ;
}
let mid = s + Math.floor((e - s) / 2);
// Buildthe 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]
function getMax(arr, index, s, e, qs, qe)
{
// If segment is not in range
if (qe < s || e < qs)
{
return Number.MIN_SAFE_INTEGER;
}
// If segment is fully in range
if (qs <= s && e <= qe) {
return seg[index];
}
// If a part of this segment is
// in range
let mid = s + Math.floor((e - s) / 2);
let left = getMax(arr, 2 * index + 1, s, mid, qs, qe);
let right = getMax(arr, 2 * index + 2, mid + 1, e, qs, qe);
// Return the maximum value
return Math.max(left, right);
}
// Driver code
let a = [5, 6, 9, 10]
let b = [3, 0, 10, -10]
let N = 4
let K = 1
maxValue(a, b, N, K);
// This code is contributed by Potta Lokesh |
4
Time Complexity: O(N*log N)
Auxiliary Space: O(N)