Search For Integers With Given Difference And At Given Distance

Posted: 4 Dec, 2020
Difficulty: Moderate

PROBLEM STATEMENT

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You are given an array consisting of 'N' non-negative integers, and two non-negative integers 'K' and 'M', your task is to find a pair of indices(say i and j) from the array such that ‘i’ ≠ ‘j’, |Ai-Aj| <= 'M', and |i-j| <= 'K', where |a-b| represents the absolute value of a-b.

Note :

1) The array may contain duplicate elements.
2) The size of the array is at least 2.
3) The given array follows 0-based indexing so 0 <= i,j< N.
4) It is guaranteed that there exist at least one such pair of indices.
Input Format :
The first line of the input contains an integer 'T' denoting the number of test cases.

The first line of each test case contains three space-separated integers 'N', 'K', and 'M', as described in the problem statement.

The second line of each test case contains 'N' space-separated integers, representing the elements of the array.
Output Format :
You just need to find the pair of indices(i and j), and if the pair returned satisfies the given condition, the output will be shown as “valid”, else the output will be shown as “invalid”.
Note :
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints :
1 <= T <= 50
1 <= N <= 10^4
1 <= K <= N
1 <= M <= 10^9
1 <= ARR[i] <= 10^9

Where 'ARR[i]' denotes the 'ith' element of the array.

Time Limit: 1 sec
Approach 1
  • Idea behind this brute force approach is to run nested loop and each time visit next k elements( because as per the problem statement the indices must have a difference of at most K) from the present index, until we find a pair that satisfies the given conditions.
  • Now, run a loop(say, loop variable i) through all the elements of the array:
  1. Run a nested loop(say, loop variable j) through the elements starting from index i+1 upto next k elements(if present, else till the end of the array).
  2. At each iteration of the inner loop check if the absolute difference of the element at index i and that at index j is at most M then return the indices i and j as the answer, else continue with the process.
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