Given two sorted array of size N. The task is to find the maximum number of elements in the first array which are strictly greater than the elements of the second array such that an element can be considered only once.
Examples:
Input: arr1[] = { 20, 30, 50 }, arr2[] = { 25, 40, 60 }
Output: 2
Explanation:
Maximum 2 elements 30 (30 > 25) and 50 (50 > 40) of array arr1 is greater than arr2.Input: arr1[] = { 10, 15, 20, 25, 30, 35 }, arr2[] = { 12, 14, 26, 32, 34, 40 }
Output: 4
Explanation:
Maximum 4 elements 15 (15 > 12), 20 (20 > 14), 30 (30 > 26) and 35 (35 > 34) of arr1 is greater than arr2.
Approach:
- Compare the elements of both the arrays from index 0 one by one.
- If the element at the index of arr1 is greater than the element at the index of arr2 then increase the answer and the index of both arrays by 1.
- If the element at the index of arr1 is lesser or equal to the element at the index of arr2 then
increase the index of arr1. - Repeat the above steps until any array’s index reaches to the last element.
- Print the answer
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find greater elements void findMaxElements(
int arr1[], int arr2[], int n)
{ // Index counter for arr1
int cnt1 = 0;
// Index counter for arr2
int cnt2 = 0;
// To store the maximum elements
int maxelements = 0;
while (cnt1 < n && cnt2 < n) {
// If element is greater,
// update maxelements and counters
// for both the arrays
if (arr1[cnt1] > arr2[cnt2]) {
maxelements++;
cnt1++;
cnt2++;
}
else {
cnt1++;
}
}
// Print the maximum elements
cout << maxelements << endl;
} int main()
{ int arr1[] = { 10, 15, 20, 25, 30, 35 };
int arr2[] = { 12, 14, 26, 32, 34, 40 };
int n = sizeof (arr1) / sizeof (arr1[0]);
findMaxElements(arr1, arr2, n);
return 0;
} |
// Java program for the above approach class Main{
// Function to find greater elements static void findmaxelements( int arr1[], int arr2[], int n)
{ // Index counter for arr1
int cnt1 = 0 ;
// Index counter for arr1
int cnt2 = 0 ;
// To store the maximum elements
int maxelements = 0 ;
while (cnt1 < n && cnt2 < n)
{
// If element is greater,
// update maxelements and counters
// for both the arrays
if (arr1[cnt1] > arr2[cnt2])
{
maxelements++;
cnt1++;
cnt2++;
}
else
{
cnt1++;
}
}
// Print the maximum elements
System.out.println(maxelements);
} // Driver Code public static void main(String[] args)
{ int arr1[] = { 10 , 15 , 20 , 25 , 30 , 35 };
int arr2[] = { 12 , 14 , 26 , 32 , 34 , 40 };
findmaxelements(arr1, arr2, arr1.length);
} } // This code is contributed by divyeshrabadiya07 |
# Python3 program for the above approach # Function to find greater elements def findmaxelements(arr1, arr2, n):
# Index counter for arr1
cnt1 = 0
# Index counter for arr2
cnt2 = 0
# To store the maximum elements
maxelements = 0 # If element is greater,
# update maxelements and counters
# for both the arrays
while cnt1 < n and cnt2 < n :
if arr1[cnt1] > arr2[cnt2] :
maxelements + = 1 cnt1 + = 1 cnt2 + = 1
else :
cnt1 + = 1
# Print the maximum elements
print (maxelements)
# Driver Code arr1 = [ 10 , 15 , 20 , 25 , 30 , 35 ]
arr2 = [ 12 , 14 , 26 , 32 , 34 , 40 ]
findmaxelements(arr1, arr2, len (arr1))
# This code is contributed by divyeshrabadiya07 |
// C# program for the above approach using System;
class GFG{
// Function to find greater elements static void findmaxelements( int [] arr1,
int [] arr2, int n)
{ // Index counter for arr1
int cnt1 = 0;
// Index counter for arr1
int cnt2 = 0;
// To store the maximum elements
int maxelements = 0;
while (cnt1 < n && cnt2 < n)
{
// If element is greater, update
// maxelements and counters for
// both the arrays
if (arr1[cnt1] > arr2[cnt2])
{
maxelements++;
cnt1++;
cnt2++;
}
else
{
cnt1++;
}
}
// Print the maximum elements
Console.Write(maxelements);
} // Driver Code static public void Main( string [] args)
{ int [] arr1 = { 10, 15, 20, 25, 30, 35 };
int [] arr2 = { 12, 14, 26, 32, 34, 40 };
findmaxelements(arr1, arr2, arr1.Length);
} } // This code is contributed by rutvik_56 |
<script> // Javascript program for the above approach // Function to find greater elements function findMaxElements( arr1, arr2, n)
{ // Index counter for arr1
var cnt1 = 0;
// Index counter for arr2
var cnt2 = 0;
// To store the maximum elements
var maxelements = 0;
while (cnt1 < n && cnt2 < n) {
// If element is greater,
// update maxelements and counters
// for both the arrays
if (arr1[cnt1] > arr2[cnt2]) {
maxelements++;
cnt1++;
cnt2++;
}
else {
cnt1++;
}
}
// Print the maximum elements
document.write( maxelements );
} var arr1 = [10, 15, 20, 25, 30, 35];
var arr2 = [12, 14, 26, 32, 34, 40];
var n = arr1.length;
findMaxElements(arr1, arr2, n); // This code is contributed by rrrtnx. </script> |
4
Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(1)