Given two unsorted arrays arr1 and arr2. They may contain duplicates. For each element in arr1 count elements less than or equal to it in array arr2.
Input : arr1 = [1, 2, 3, 4, 7, 9] arr2 = [0, 1, 2, 1, 1, 4] Output : [4, 5, 5, 6, 6, 6] Input : arr1 = [5, 10, 2, 6, 1, 8, 6, 12] arr2 = [6, 5, 11, 4, 2, 3, 7] Output : [4, 6, 1, 5, 0, 6, 5, 7]
This problem is already discussed in the previous post. In this article, a more optimized linear time solution to the above problem is discussed. The approach discussed here works for arrays with values in small range. A range of values that can be used as index in an array.
The idea is to use an array to create a direct address table initially filled with zero, such that hash[i] gives the count of an element in the second array arr2. Now, calculate pre-sum of the hash array. Doing this, hash[i] will now give the count of elements less than or equal to in second array arr2.
Now, traverse the first array and print hash[arr1[i]].
Below program illustrate the above approach:
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Another method:(Using vector and map)
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- For each element in 1st array count elements less than or equal to it in 2nd array
- Count equal element pairs in the given array
- Check if the array has an element which is equal to sum of all the remaining elements
- Check if the array has an element which is equal to XOR of remaining elements
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- Count pairs in an array such that both elements has equal set bits
- Count of index pairs with equal elements in an array
- Count of smaller or equal elements in sorted array
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- Count of triplets from the given Array such that sum of any two elements is the third element
- Count elements that are divisible by at-least one element in another array
- Find an element in array such that sum of left array is equal to sum of right array
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