Javascript Program to Maximize count of corresponding same elements in given Arrays by Rotation
Last Updated :
04 Feb, 2022
Given two arrays arr1[] and arr2[] of N integers and array arr1[] has distinct elements. The task is to find the maximum count of corresponding same elements in the given arrays by performing cyclic left or right shift on array arr1[].
Examples:
Input: arr1[] = { 6, 7, 3, 9, 5 }, arr2[] = { 7, 3, 9, 5, 6 }
Output: 5
Explanation:
By performing cyclic left shift on array arr1[] by 1.
Updated array arr1[] = {7, 3, 9, 5, 6}.
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].
Input: arr1[] = {1, 3, 2, 4}, arr2[] = {4, 2, 3, 1}
Output: 2
Explanation:
By performing cyclic left shift on array arr1[] by 1.
Updated array arr1[] = {3, 2, 4, 1}
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].
Approach: This problem can be solved using Greedy Approach. Below are the steps:
- Store the position of all the elements of the array arr2[] in an array(say store[]).
- For each element in the array arr1[], do the following:
- Find the difference(say diff) between the position of the current element in arr2[] with the position in arr1[].
- If diff is less than 0 then update diff to (N – diff).
- Store the frequency of current difference diff in a map.
- After the above steps, the maximum frequency stored in map is the maximum number of equal elements after rotation on arr1[].
Below is the implementation of the above approach:
Javascript
<script>
function maximumEqual(a, b, n)
{
let store = Array.from({length: 1e5}, (_, i) => 0);
for (let i = 0; i < n; i++)
{
store[b[i]] = i + 1;
}
let ans = Array.from({length: 1e5}, (_, i) => 0);
for (let i = 0; i < n; i++)
{
let d = Math.abs(store[a[i]] - (i + 1));
if (store[a[i]] < i + 1)
{
d = n - d;
}
ans[d]++;
}
let finalans = 0;
for (let i = 0; i < 1e5; i++)
finalans = Math.max(finalans,
ans[i]);
document.write(finalans + "
" );
}
let A = [ 6, 7, 3, 9, 5 ];
let B = [ 7, 3, 9, 5, 6 ];
let size = A.length;
maximumEqual(A, B, size);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Please refer complete article on Maximize count of corresponding same elements in given Arrays by Rotation for more details!
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