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How you can improve search efficiency of sequential search for a sorted file. Discuss limitations in such?
Wiki User
∙ 6y ago
When sequentially searching n items, the best-case is O(1) and the worst-case is O(n). But when the items are sorted, binary search will improve efficiency. The best case is still O(1), but worst case drops to O(log n) where log n is the binary logarithm of n.
Binary search starts with the middle element of the set. If the set is empty, the item we're looking for does not exist but if the middle element is the item we are looking for then we are done. If not, a simple comparison will tell us in which half of the set to discard (including the middle element). We repeat the process with the remaining half. If there are no elements remaining, the item does not exist.
Wiki User
∙ 6y ago
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When sequentially searching n items, the best-case is O(1) and the worst-case is O(n). But when the items are sorted, binary search will improve efficiency. The best case is still O(1), but worst case drops to O(log n) where log n is the binary logarithm of n. Binary search starts with the middle element of the set. If the set is empty, the item we're looking for does not exist but if the middle element is the item we are looking for then we are done. If not, a simple comparison will tell us in which half of the set to discard (including the middle element). We repeat the process with the remaining half. If there are no elements remaining, the item does not exist.
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How you can improve search efficiency of sequential search for a sorted file. Discuss limitations in such.?
When sequentially searching n items, the best-case is O(1) and the worst-case is O(n). But when the items are sorted, binary search will improve efficiency. The best case is still O(1), but worst case drops to O(log n) where log n is the binary logarithm of n. Binary search starts with the middle element of the set. If the set is empty, the item we're looking for does not exist but if the middle element is the item we are looking for then we are done. If not, a simple comparison will tell us in which half of the set to discard (including the middle element). We repeat the process with the remaining half. If there are no elements remaining, the item does not exist.
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