Given an array of n integers. Find the maximum value of arr[i] mod arr[j] where arr[i] >= arr[j] and 1 <= i, j <= n
Examples:
Input: arr[] = {3, 4, 7}
Output: 3
Explanation:
There are 3 pairs which satisfies arr[i] >= arr[j] are:-
4, 3 => 4 % 3 = 1
7, 3 => 7 % 3 = 1
7, 4 => 7 % 4 = 3
Hence Maximum value among all is 3.
Input: arr[] = {3, 7, 4, 11}
Output: 4
Input: arr[] = {4, 4, 4}
Output: 0
A Naive approach is to run two nested for loops and select the maximum of every possible pairs after taking modulo of them. Time complexity of this approach will be O(n2) which will not be sufficient for large value of n.
An efficient approach (when elements are from small range) is to use sorting and binary search method. Firstly we will sort the array so that we would able to apply binary search on it. Since we need to maximize the value of arr[i] mod arr[j] so we iterate through each x(such x divisible by arr[j]) in range from 2*arr[j] to M+arr[j], where M is Maximum value of sequence. For each value of x we need to find maximum value of arr[i] such that arr[i] < x.
By doing this we would assure that we have chosen only those values of arr[i] that will give the maximum value of arr[i] mod arr[j]. After that we just need to repeat the above process for other values of arr[j] and update the answer by value a[i] mod arr[j]. For example:-
If arr[] = {4, 6, 7, 8, 10, 12, 15} then for
first element, i.e., arr[j] = 4 we iterate
through x = {8, 12, 16}.
Therefore for each value of x, a[i] will be:-
x = 8, arr[i] = 7 (7 < 8)
ans = 7 mod 4 = 3
x = 12, arr[i] = 10 (10 < 12)
ans = 10 mod 4 = 2 (Since 2 < 3,
No update)
x = 16, arr[i] = 15 (15 < 16)
ans = 15 mod 4 = 3 (Since 3 == 3,
No need to update)
C++
// C++ program to find Maximum modulo value #include <bits/stdc++.h> using namespace std; int maxModValue(int arr[], int n) { int ans = 0; // Sort the array[] by using inbuilt sort function sort(arr, arr + n); for (int j = n - 2; j >= 0; --j) { // Break loop if answer is greater or equals to // the arr[j] as any number modulo with arr[j] // can only give maximum value up-to arr[j]-1 if (ans >= arr[j]) break; // If both elements are same then skip the next // loop as it would be worthless to repeat the // rest process for same value if (arr[j] == arr[j + 1]) continue; for (int i = 2 * arr[j]; i <= arr[n - 1] + arr[j]; i += arr[j]) { // Fetch the index which is greater than or // equals to arr[i] by using binary search // inbuilt lower_bound() function of C++ int ind = lower_bound(arr, arr + n, i) - arr; // Update the answer ans = max(ans, arr[ind - 1] % arr[j]); } } return ans; } // Driver code int main() { int arr[] = { 3, 4, 5, 9, 11 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxModValue(arr, n); } |
Java
// Java program to find Maximum modulo value import java.util.Arrays; class Test { static int maxModValue(int arr[], int n) { int ans = 0; // Sort the array[] by using inbuilt sort function Arrays.sort(arr); for (int j = n - 2; j >= 0; --j) { // Break loop if answer is greater or equals to // the arr[j] as any number modulo with arr[j] // can only give maximum value up-to arr[j]-1 if (ans >= arr[j]) break; // If both elements are same then skip the next // loop as it would be worthless to repeat the // rest process for same value if (arr[j] == arr[j + 1]) continue; for (int i = 2 * arr[j]; i <= arr[n - 1] + arr[j]; i += arr[j]) { // Fetch the index which is greater than or // equals to arr[i] by using binary search int ind = Arrays.binarySearch(arr, i); if (ind < 0) ind = Math.abs(ind + 1); else { while (arr[ind] == i) { ind--; if (ind == 0) { ind = -1; break; } } ind++; } // Update the answer ans = Math.max(ans, arr[ind - 1] % arr[j]); } } return ans; } // Driver method public static void main(String args[]) { int arr[] = { 3, 4, 5, 9, 11 }; System.out.println(maxModValue(arr, arr.length)); } } |
Python3
# Python3 program to find Maximum modulo value def maxModValue(arr, n): ans = 0 # Sort the array[] by using inbuilt # sort function arr = sorted(arr) for j in range(n - 2, -1, -1): # Break loop if answer is greater or equals to # the arr[j] as any number modulo with arr[j] # can only give maximum value up-to arr[j]-1 if (ans >= arr[j]): break # If both elements are same then skip the next # loop as it would be worthless to repeat the # rest process for same value if (arr[j] == arr[j + 1]) : continue i = 2 * arr[j] while(i <= arr[n - 1] + arr[j]): # Fetch the index which is greater than or # equals to arr[i] by using binary search # inbuilt lower_bound() function of C++ ind = 0 for k in arr: if k >= i: ind = arr.index(k) # Update the answer ans = max(ans, arr[ind - 1] % arr[j]) i += arr[j] return ans # Driver Code arr = [3, 4, 5, 9, 11 ] n = 5print(maxModValue(arr, n)) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to find Maximum modulo value using System; public class GFG { static int maxModValue(int[] arr, int n) { int ans = 0; // Sort the array[] by using inbuilt // sort function Array.Sort(arr); for (int j = n - 2; j >= 0; --j) { // Break loop if answer is greater // or equals to the arr[j] as any // number modulo with arr[j] can // only give maximum value up-to // arr[j]-1 if (ans >= arr[j]) break; // If both elements are same then // skip the next loop as it would // be worthless to repeat the // rest process for same value if (arr[j] == arr[j + 1]) continue; for (int i = 2 * arr[j]; i <= arr[n - 1] + arr[j]; i += arr[j]) { // Fetch the index which is // greater than or equals to // arr[i] by using binary search int ind = Array.BinarySearch(arr, i); if (ind < 0) ind = Math.Abs(ind + 1); else { while (arr[ind] == i) { ind--; if (ind == 0) { ind = -1; break; } } ind++; } // Update the answer ans = Math.Max(ans, arr[ind - 1] % arr[j]); } } return ans; } // Driver method public static void Main() { int[] arr = { 3, 4, 5, 9, 11 }; Console.WriteLine( maxModValue(arr, arr.Length)); } } // This code is contributed by Sam007. |
Output: 4
Time complexity: O(nlog(n) + Mlog(M)) where n is total number of elements and M is maximum value of all the elements.
Auxiliary space: O(1)
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