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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum// remainder of sum of a pair// of array elements modulo Kvoid maxRemainder(int A[], int N, int K){ // Stores all even numbers unordered_set<int> even; // Stores all odd numbers set<int> odd; // Segregate remainders of even // and odd numbers in respective sets for(int i = 0; i < N; i++) { int num = A[i]; if (num % 2 == 0) even.insert(num % K); else odd.insert(num % K); } // Stores the maximum // remainder obtained int max_rem = 0; // Find the complement of remainder // of each even number in odd set for(int x : even) { // Find the complement // of remainder x int y = K - 1 - x; auto it = odd.upper_bound(y); if (it != odd.begin()) { it--; max_rem = max(max_rem, x + *it); } } // Print the answer cout << max_rem;}// Driver codeint main(){ // Given array int arr[] = { 3, 2, 4, 11, 6, 7 }; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); // Given value of K int K = 7; maxRemainder(arr, N, K); return 0;}// This code is contributed by Kingash |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to find the maximum // remainder of sum of a pair // of array elements modulo K static void maxRemainder(int A[], int N, int K) { // Stores all even numbers HashSet<Integer> even = new HashSet<>(); // Stores all odd numbers TreeSet<Integer> odd = new TreeSet<>(); // Segregate remainders of even // and odd numbers in respective sets for (int num : A) { if (num % 2 == 0) even.add(num % K); else odd.add(num % K); } // Stores the maximum // remainder obtained int max_rem = 0; // Find the complement of remainder // of each even number in odd set for (int x : even) { // Find the complement // of remainder x int y = K - 1 - x; if (odd.floor(y) != null) max_rem = Math.max( max_rem, x + odd.floor(y)); } // Print the answer System.out.print(max_rem); } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 3, 2, 4, 11, 6, 7 }; // Size of the array int N = arr.length; // Given value of K int K = 7; maxRemainder(arr, N, K); }} |
Python3
# Python3 program for the above approachfrom bisect import bisect_left# Function to find the maximum# remainder of sum of a pair# of array elements modulo Kdef maxRemainder(A, N, K): # Stores all even numbers even = {} # Stores all odd numbers odd = {} # Segregate remainders of even # and odd numbers in respective sets for i in range(N): num = A[i] if (num % 2 == 0): even[num % K] = 1 else: odd[num % K] = 1 # Stores the maximum # remainder obtained max_rem = 0 # Find the complement of remainder # of each even number in odd set for x in even: # Find the complement # of remainder x 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 the answer print (max_rem)# Driver codeif __name__ == '__main__': # Given array arr = [3, 2, 4, 11, 6, 7] # Size of the array N = len(arr) # Given value of K K = 7 maxRemainder(arr, N, K) # This code is contributed by mohit kumar 29 |
C#
using System;using System.Collections.Generic;class GFG{ // Function to find the maximum // remainder of sum of a pair // of array elements modulo K static void MaxRemainder(int[] A, int N, int K) { // Stores all even numbers HashSet<int> even = new HashSet<int>(); // Stores all odd numbers SortedSet<int> odd = new SortedSet<int>(); // Segregate remainders of even // and odd numbers in respective sets foreach (int num in A) { if (num % 2 == 0) even.Add(num % K); else odd.Add(num % K); } // Stores the maximum // remainder obtained int max_rem = 0; // Find the complement of remainder // of each even number in odd set foreach (int x in even) { // Find the complement // of remainder x int y = K - 1 - x; if (odd.Min <= y) max_rem = Math.Max(max_rem, x + (int)odd.GetViewBetween(int.MinValue, y).Max); } // Print the answer Console.Write(max_rem); } // Driver Code public static void Main(string[] args) { // Given array int[] arr = { 3, 2, 4, 11, 6, 7 }; // Size of the array int N = arr.Length; // Given value of K int K = 7; MaxRemainder(arr, N, K); }}// This code is contributed by phasing17 |
Javascript
// JavaScript program for the above approach// Function to find the upper bound of an array. function upper_bound(arr, X){ let mid; // Initialise starting index and // ending index let low = 0; let high = arr.length; // Till low is less than high while (low < high) { // Find the middle index mid = low + Math.floor((high - low) / 2); // If X is greater than or equal // to arr[mid] then find // in right subarray if (X >= arr[mid]) { low = mid + 1; } // If X is less than arr[mid] // then find in left subarray else { high = mid; } } // if X is greater than arr[n-1] if(low < N && arr[low] <= X) { low++; } // Return the upper_bound index return low;}// Function to find the maximum// remainder of sum of a pair// of array elements modulo Kfunction maxRemainder(A, N, K){ // Stores all even numbers let even = new Array(); // Stores all odd numbers let odd = new Array(); // Segregate remainders of even // and odd numbers in respective sets 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(); // Stores the maximum // remainder obtained let max_rem = 0; // Find the complement of remainder // of each even number in odd set for(let i = 0;i < even.length; i++){ let x = even[i]; // Find the complement // of remainder xd // console.log(x); let y = K - 1 - x; let it = upper_bound(odd, y); if (it != 0) { it--; max_rem = Math.max(max_rem, x + odd[it]); } } // Print the answer console.log(max_rem);}// Driver code// Given arraylet arr = [3, 2, 4, 11, 6, 7];// Size of the arraylet N = arr.length;// Given value of Klet K = 7;maxRemainder(arr, N, K);// This code is contributed by Nidhi goel |
Output:
6
Time Complexity: O(N * logN)
Auxiliary Space: O(N)
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