Maximize remainder of sum of a pair of array elements with different parity modulo K
Given an array arr[] of size N, consisting of N / 2 even and odd integers each, and an integer K, the task is to find the maximum remainder of sum of a pair of array elements of different parity modulo K.
Examples:
Input: arr[] = {3, 2, 4, 11, 6, 7}, K = 7
Output: 6
Explanation:
Sum of a pair of array elements = 2 + 11
Sum % K = 13 % 7 = 6.
Therefore, the maximum remainder possible is 6.
Input: arr[] = {8, 11, 17, 16}, K = 13
Output: 12
Approach: Follow the steps below to solve the problem:
- Initialize a HashSet, say even, to store all even array elements.
- Initialize a TreeSet, say odd, to store all odd array elements.
- Initialize a variable, say max_rem, to store the maximum remainder possible.
- Traverse the HashSet and for each element, find its complement and search for it in the set odd, which is less than equal to its complement.
- Update max_rem with the sum of elements, and it’s complement.
- Print the maximum remainder i.e. value of max_rem.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void maxRemainder( int A[], int N, int K)
{
unordered_set< int > even;
set< int > odd;
for ( int i = 0; i < N; i++)
{
int num = A[i];
if (num % 2 == 0)
even.insert(num % K);
else
odd.insert(num % K);
}
int max_rem = 0;
for ( int x : even)
{
int y = K - 1 - x;
auto it = odd.upper_bound(y);
if (it != odd.begin())
{
it--;
max_rem = max(max_rem, x + *it);
}
}
cout << max_rem;
}
int main()
{
int arr[] = { 3, 2, 4, 11, 6, 7 };
int N = sizeof (arr) / sizeof (arr[0]);
int K = 7;
maxRemainder(arr, N, K);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void maxRemainder( int A[],
int N, int K)
{
HashSet<Integer> even
= new HashSet<>();
TreeSet<Integer> odd
= new TreeSet<>();
for ( int num : A) {
if (num % 2 == 0 )
even.add(num % K);
else
odd.add(num % K);
}
int max_rem = 0 ;
for ( int x : even) {
int y = K - 1 - x;
if (odd.floor(y) != null )
max_rem
= Math.max(
max_rem,
x + odd.floor(y));
}
System.out.print(max_rem);
}
public static void main(String[] args)
{
int arr[] = { 3 , 2 , 4 , 11 , 6 , 7 };
int N = arr.length;
int K = 7 ;
maxRemainder(arr, N, K);
}
}
|
Python3
from bisect import bisect_left
def maxRemainder(A, N, K):
even = {}
odd = {}
for i in range (N):
num = A[i]
if (num % 2 = = 0 ):
even[num % K] = 1
else :
odd[num % K] = 1
max_rem = 0
for x in even:
y = K - 1 - x
od = list (odd.keys())
it = bisect_left(od, y)
if (it ! = 0 ):
max_rem = max (max_rem, x + od[it])
print (max_rem)
if __name__ = = '__main__' :
arr = [ 3 , 2 , 4 , 11 , 6 , 7 ]
N = len (arr)
K = 7
maxRemainder(arr, N, K)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void MaxRemainder( int [] A, int N, int K)
{
HashSet< int > even = new HashSet< int >();
SortedSet< int > odd = new SortedSet< int >();
foreach ( int num in A)
{
if (num % 2 == 0)
even.Add(num % K);
else
odd.Add(num % K);
}
int max_rem = 0;
foreach ( int x in even)
{
int y = K - 1 - x;
if (odd.Min <= y)
max_rem = Math.Max(max_rem, x + ( int )odd.GetViewBetween( int .MinValue, y).Max);
}
Console.Write(max_rem);
}
public static void Main( string [] args)
{
int [] arr = { 3, 2, 4, 11, 6, 7 };
int N = arr.Length;
int K = 7;
MaxRemainder(arr, N, K);
}
}
|
Javascript
function upper_bound(arr, X)
{
let mid;
let low = 0;
let high = arr.length;
while (low < high) {
mid = low + Math.floor((high - low) / 2);
if (X >= arr[mid]) {
low = mid + 1;
}
else {
high = mid;
}
}
if (low < N && arr[low] <= X) {
low++;
}
return low;
}
function maxRemainder(A, N, K)
{
let even = new Array();
let odd = new Array();
for (let i = 0; i < N; i++)
{
let num = A[i];
if (num % 2 == 0 && !even.includes(num%K)){
even.push(num % K);
}
else if (num%2 != 0 && !odd.includes(num%K)){
odd.push(num % K);
}
}
odd.sort();
let max_rem = 0;
for (let i = 0;i < even.length; i++){
let x = even[i];
let y = K - 1 - x;
let it = upper_bound(odd, y);
if (it != 0)
{
it--;
max_rem = Math.max(max_rem, x + odd[it]);
}
}
console.log(max_rem);
}
let arr = [3, 2, 4, 11, 6, 7];
let N = arr.length;
let K = 7;
maxRemainder(arr, N, K);
|
Time Complexity: O(N * logN)
Auxiliary Space: O(N)
Last Updated :
10 Jan, 2023
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