kotlin-coding-challenges

Binary search

Instructions

Given list of sorted elements and a element return index of that element in the list or -1 if element was not found. Use binary search algorithm.

Challenge

Solution

Algorithm

We ignore half of the elements after each loop.

• Compare element with the middle element.
• If element matches with middle element, we return the middle element index.
• Else If element is greater than the mid element, then element can only lie in right half sub-list after the mid element. So we deal only with right half.
• Else (x is smaller) deal only with the left half.

Examples

Example 1

Search for C in [A, B, C, D, E, F, G, H, I, J, K, L, M, O, P]

[A, B, C, D, E, F, G, H, I, J, K, L, M, O, P] // (left = A, middle = H, right = P)
[A, B, C, D, E, F, G, H, I, J, K, L, M, O, P] // (left = A, middle = D, right = G)
[A, B, C, D, E, F, G, H, I, J, K, L, M, O, P] // (left = A, middle = C, right = F)

Example 2

binarySearch(listOf('A', 'B', 'C'), 'A') // 0

binarySearch(listOf('A', 'B', 'C'), 'B') // 1

binarySearch(listOf('A', 'B', 'C'), 'H') // -1

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