You are given an array of n-elements with a basic condition that occurrence of greatest element is more than once. You have to find the minimum distance between maximums. (n>=2).
Examples:
Input : arr[] = {3, 5, 2, 3, 5, 3, 5}
Output : Minimum Distance = 2
Explanation : Greatest element is 5 and its index
are 1, 4 and 6. Resulting minimum distance of 2
from position 4 to 6.
Input : arr[] = {1, 1, 1, 1, 1, 1}
Output : Minimum Distance = 1
Explanation : Greatest element is 1 and its index
are 0, 1, 2, 3, 4 and 5. Resulting minimum distance
of 1.
A basic approach runs in O(n2). First we find the maximum element. Then for each element equal to maximum element, we find the nearest maximum element.
An efficient solution finish our job in a single traversing of array. We initialize maximum_element = arr[0], min_distance = n and index = 0. After that for each element we should check whether element is equal to, greater or less than maximum element. Depending upon three cases we have following option:
- case a: If element is equal to maximum_element then update min_dis = min( min_dis , (i-index)) and update index = i;
- case b: If element is greater than maximum_element update maximum_element = arr[i], index = i and min_dis = n.
- case c: If element is less than maximum_element then iterate to next element.
Implementation:
C++
// C++ program to find Min distance // of maximum element#include<bits/stdc++.h>using namespace std;//function to return min distanceint minDistance (int arr[], int n){ int maximum_element = arr[0]; int min_dis = n; int index = 0; for (int i=1; i<n; i++) { // case a if (maximum_element == arr[i]) { min_dis = min(min_dis, (i-index)); index = i; } // case b else if (maximum_element < arr[i]) { maximum_element = arr[i]; min_dis = n; index = i; } // case c else continue; } return min_dis;}// driver programint main(){ int arr[] = {6, 3, 1, 3, 6, 4, 6}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Minimum distance = " << minDistance(arr, n); return 0;} |
Java
// Java program to find Min distance // of maximum elementclass GFG { // function to return min distance static int minDistance (int arr[], int n) { int maximum_element = arr[0]; int min_dis = n; int index = 0; for (int i=1; i<n; i++) { // case a if (maximum_element == arr[i]) { min_dis = Math.min(min_dis, (i-index)); index = i; } // case b else if (maximum_element < arr[i]) { maximum_element = arr[i]; min_dis = n; index = i; } // case c else continue; } return min_dis; } // Driver code public static void main (String[] args) { int arr[] = {6, 3, 1, 3, 6, 4, 6}; int n = arr.length; System.out.print("Minimum distance = "+minDistance(arr, n)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find Min # distance of maximum element# Function to return min distancedef minDistance (arr, n): maximum_element = arr[0] min_dis = n index = 0 for i in range(1, n): # case a if (maximum_element == arr[i]): min_dis = min(min_dis, (i - index)) index = i # case b elif (maximum_element < arr[i]): maximum_element = arr[i] min_dis = n index = i # case c else: continue return min_dis# driver programarr = [6, 3, 1, 3, 6, 4, 6]n = len(arr)print("Minimum distance =", minDistance(arr, n)) # This code is contributed by Smitha Dinesh Semwal |
C#
// C# program to find Min distance // of maximum elementusing System;class GFG { // function to return min distance static int minDistance (int []arr, int n) { int maximum_element = arr[0]; int min_dis = n; int index = 0; for (int i = 1; i < n; i++) { // case a if (maximum_element == arr[i]) { min_dis = Math.Min(min_dis, (i - index)); index = i; } // case b else if (maximum_element < arr[i]) { maximum_element = arr[i]; min_dis = n; index = i; } // case c else continue; } return min_dis; } // Driver code public static void Main () { int []arr = {6, 3, 1, 3, 6, 4, 6}; int n = arr.Length; Console.WriteLine("Minimum distance = " + minDistance(arr, n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to find Min distance // of maximum element//function to return min distancefunction minDistance($arr, $n){ $maximum_element = $arr[0]; $min_dis = $n; $index = 0; for ($i = 1; $i < $n; $i++) { // case a if ($maximum_element == $arr[$i]) { $min_dis = min($min_dis, ($i - $index)); $index = $i; } // case b else if ($maximum_element < $arr[$i]) { $maximum_element = $arr[$i]; $min_dis = $n; $index = $i; } // case c else continue; }return $min_dis;} // Driver Code $arr = array(6, 3, 1, 3, 6, 4, 6); $n = count($arr); echo "Minimum distance = " .minDistance($arr,$n); // This code is contributed by Sam007?> |
Javascript
<script>// JavaScript program to find Min distance // of maximum element // function to return min distance function minDistance (arr, n) { let maximum_element = arr[0]; let min_dis = n; let index = 0; for (let i=1; i<n; i++) { // case a if (maximum_element == arr[i]) { min_dis = Math.min(min_dis, (i-index)); index = i; } // case b else if (maximum_element < arr[i]) { maximum_element = arr[i]; min_dis = n; index = i; } // case c else continue; } return min_dis; } // Driver code let arr = [6, 3, 1, 3, 6, 4, 6]; let n = arr.length; document.write("Minimum distance = "+minDistance(arr, n));</script> |
Output
Minimum distance = 2
Time Complexity: O(n)
Auxiliary Space: O(1)
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