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vishal chugh
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in DS
Jan 24, 2018

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What is the worst case time complexity to find *k*^{th} smallest element into an array of ‘*n*’ element?

@vishal chugh,look why it will take o(n) in worst case because here the comparison will be done with every element of the list. This is the case mostly in linear search. However, if you go to see the case of Binary Search,here the prerequisite only is that the list should be sorted. This will take less time in comparison to the linear search. :)

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Following is complete algorithm.

kthSmallest(arr[0..n-1], k)

1)Divide arr[] into ⌈n/5⌉ groups where size of each group is 5 except possibly the last group which may have less than 5 elements.

2)Sort the above created ⌈n/5⌉ groups and find median of all groups. Create an auxiliary array ‘median[]’ and store medians of all ⌈n/5⌉ groups in this median array.// Recursively call this method to find median of median[0..⌈n/5⌉-1]

3)medOfMed = kthSmallest(median[0..⌈n/5⌉-1], ⌈n/10⌉)

4)Partition arr[] around medOfMed and obtain its position.

pos = partition(arr, n, medOfMed)

5)If pos == k return medOfMed

6)If pos > k return kthSmallest(arr[l..pos-1], k)

7)If pos < k return kthSmallest(arr[pos+1..r], k-pos+l-1)

// C++ implementation of worst case linear time algorithm // to find k'th smallest element #include<iostream> #include<algorithm> #include<climits> using namespace std; int partition(int arr[], int l, int r, int k); // A simple function to find median of arr[]. This is called // only for an array of size 5 in this program. int findMedian(int arr[], int n) { sort(arr, arr+n); // Sort the array return arr[n/2]; // Return middle element } // Returns k'th smallest element in arr[l..r] in worst case // linear time. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT int kthSmallest(int arr[], int l, int r, int k) { // If k is smaller than number of elements in array if (k > 0 && k <= r - l + 1) { int n = r-l+1; // Number of elements in arr[l..r] // Divide arr[] in groups of size 5, calculate median // of every group and store it in median[] array. int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups; for (i=0; i<n/5; i++) median[i] = findMedian(arr+l+i*5, 5); if (i*5 < n) //For last group with less than 5 elements { median[i] = findMedian(arr+l+i*5, n%5); i++; } // Find median of all medians using recursive call. // If median[] has only one element, then no need // of recursive call int medOfMed = (i == 1)? median[i-1]: kthSmallest(median, 0, i-1, i/2); // Partition the array around a random element and // get position of pivot element in sorted array int pos = partition(arr, l, r, medOfMed); // If position is same as k if (pos-l == k-1) return arr[pos]; if (pos-l > k-1) // If position is more, recur for left return kthSmallest(arr, l, pos-1, k); // Else recur for right subarray return kthSmallest(arr, pos+1, r, k-pos+l-1); } // If k is more than number of elements in array return INT_MAX; } void swap(int *a, int *b) { int temp = *a; *a = *b; *b = temp; } // It searches for x in arr[l..r], and partitions the array // around x. int partition(int arr[], int l, int r, int x) { // Search for x in arr[l..r] and move it to end int i; for (i=l; i<r; i++) if (arr[i] == x) break; swap(&arr[i], &arr[r]); // Standard partition algorithm i = l; for (int j = l; j <= r - 1; j++) { if (arr[j] <= x) { swap(&arr[i], &arr[j]); i++; } } swap(&arr[i], &arr[r]); return i; } // Driver program to test above methods int main() { int arr[] = {12, 3, 5, 7, 4, 19, 26}; int n = sizeof(arr)/sizeof(arr[0]), k = 3; cout << "K'th smallest element is " << kthSmallest(arr, 0, n-1, k); return 0; }

**Time Complexity:**

The worst case time complexity of the above algorithm is O(n). Let us analyze all steps.