48 lines
1.3 KiB
Python
48 lines
1.3 KiB
Python
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"""
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TERNARY SEARCH
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Time Complexity: O(log3 N)
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Working: Similar to binary search but here we have two searc point (mid)
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Steps:
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1. Compare with elements at mid1 and mid2. Stop if found
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2. If more than mid1 and less than mid2 - working field is the second part
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3. If less than mid1 - working field is the first part
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4. If more than mid2 - working field is the third part
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"""
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arr = [10, 13, 46, 51, 139, 467]
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targets = [2, 467]
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def ternarySearch(arr, target):
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left, right = 0, len(arr) - 1
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index = -1
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while(left <= right):
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mid1, mid2 = left + (right - left) // 3, right - (right - left) // 3
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# checking for condition 1
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if target == arr[mid1]:
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index = mid1
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break
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if target == arr[mid2]:
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index = mid2
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break
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# checking for condition 2
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if target > arr[mid1] and target < arr[mid2]:
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left, right = mid1 + 1, mid2 - 1
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# checking for condition 3
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elif target < arr[mid1]:
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right = mid1 - 1
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# checking for condition 4
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else:
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left = mid2 + 1
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return index
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for target in targets:
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result = ternarySearch(arr, target)
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if result == -1:
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print(target, "is not present in", arr)
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else:
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print(target, "is present in the position:", result + 1)
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