|
| 1 | +import heapq |
| 2 | + |
| 3 | +def find_kth_largest(nums, k): |
| 4 | + |
| 5 | + # Initialize an empty list |
| 6 | + k_numbers_min_heap = [] |
| 7 | + |
| 8 | + # Add first k elements to the list |
| 9 | + for i in range(k): |
| 10 | + k_numbers_min_heap.append(nums[i]) |
| 11 | + |
| 12 | + # Convert list to min_heap |
| 13 | + heapq.heapify(k_numbers_min_heap) |
| 14 | + |
| 15 | + # Iterate through remaining elements in num array |
| 16 | + for i in range(k, len(nums)): |
| 17 | + |
| 18 | + # Compare current element with min heap element |
| 19 | + if nums[i] > k_numbers_min_heap[0]: |
| 20 | + |
| 21 | + # Remove smallest element |
| 22 | + heapq.heappop(k_numbers_min_heap) |
| 23 | + |
| 24 | + # Add current element |
| 25 | + heapq.heappush(k_numbers_min_heap, nums[i]) |
| 26 | + |
| 27 | + return k_numbers_min_heap[0] |
| 28 | + |
| 29 | + |
| 30 | + |
| 31 | +# Time Complexity = O((n-k)logk) |
| 32 | +# Space Complexity = O(k) |
| 33 | + |
| 34 | + |
| 35 | + |
| 36 | +############################################################################ |
| 37 | + |
| 38 | + |
| 39 | + |
| 40 | +# Driver code |
| 41 | +def main(): |
| 42 | + input_array = [ |
| 43 | + [1, 5, 12, 2, 11, 9, 7, 30, 20], |
| 44 | + [5, 2, 9, -3, 7], |
| 45 | + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], |
| 46 | + [1, 4, 6, 0, 2], |
| 47 | + [3, 5, 2, 3, 8, 5, 3] |
| 48 | + ] |
| 49 | + |
| 50 | + k = [3, 1, 9, 1, 4] |
| 51 | + |
| 52 | + for i in range(len(input_array)): |
| 53 | + print(i + 1, ".\tInput array: ", input_array[i], sep="") |
| 54 | + print("\tValue of k: ", k[i], sep="") |
| 55 | + print("\tkth largest number: ", find_kth_largest(input_array[i], k[i]), sep="") |
| 56 | + print("-" * 100) |
| 57 | + |
| 58 | + |
| 59 | +if __name__ == '__main__': |
| 60 | + main() |
| 61 | + |
| 62 | + |
| 63 | + |
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