Count number of elements between two given elements in array
Given an unsorted array of n elements and also given two points num1 and num2. The task is to count number of elements occurs between the given points (excluding num1 and num2).
If there are multiple occurrences of num1 and num2, we need to consider leftmost occurrence of num1 and rightmost occurrence of num2.
Examples:
Input : arr[] = {3 5 7 6 4 9 12 4 8} num1 = 5 num2 = 4 Output : 5 Number of elements between leftmost occurrence of 5 and rightmost occurrence of 4 is five. Input : arr[] = {4, 6, 8, 3, 6, 2, 8, 9, 4} num1 = 4 num2 = 4 Output : 7 Input : arr[] = {4, 6, 8, 3, 6, 2, 8, 9, 4} num1 = 4 num2 = 10 Output : 0
The solution should traverse array only once in all cases (when single or both elements are not present)
The idea is to traverse array from left and find first occurrence of num1. If we reach end, we return 0. Then we traverse from rightmost element and find num2. We traverse only till the point which is greater than index of num1. If we reach end, we return 0. If we found both elements, we return count using indexes of found elements.
Implementation:
CPP
Java
Python3
C#
PHP
Javascript
<script> // Program to count number of elements between // two given elements. // Function to count number of elements // occurs between the elements. function getCount( arr, n, num1, num2) { // Find num1 let i = 0; for (i = 0; i < n; i++) if (arr[i] == num1) break ; // If num1 is not present or present at end if (i >= n-1) return 0; // Find num2 let j; for (j = n-1; j >= i+1; j--) if (arr[j] == num2) break ; // If num2 is not present if (j == i) return 0; // return number of elements between // the two elements. return (j - i - 1); } // Driver program let arr = [ 3, 5, 7, 6, 4, 9, 12, 4, 8 ]; let n = arr.length; let num1 = 5, num2 = 4; document.write(getCount(arr, n, num1, num2)); </script> |
5
Time Complexity: O(n)
Auxiliary Space: O(1)
How to handle multiple queries?
For handling multiple queries, we can use hashing and store leftmost and rightmost indexes for every element present in array. Once we have stored these, we can answer all queries in O(1) time.
This article is contributed by Dharmendra kumar.
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