Woodstock Blog

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[Fundamental] Quickselect



Find Top k smallest element in an array.


There’re 2 solutions.

First solution, use a max-heap. O(nlgk) time complexity.

Second solution is called quickselect, a type of selection algorithm that’s based on quicksort. It’s averaging O(n) time, but O(n2) if pivot selection is poor. The code is posted below. There’s also a similar iterative solution.

To further optimize this, we can change the pivot selection method by dividing into k group and find median of each. This is called Median of medians algorithm. The worst case is O(n) time. And this is the best solution for “Top k” questions.

Why quickselect is O(n) time?

It’s a very good question to ask. Why O(n)?

Well think about it. Let’s assume you always find the pivot that makes you eliminate half of the input.

The first run, you would read n elements. Second time you read half of n, and third time, quarter of n. In the end, you read n + n/2 + n/4 + … = 2n times.

Compared to the Heap method to find top K, quick select has its advantage. Heap top K take O(n lgK) time. So when K is pretty large, quick select might be an better solution.



public static void quickSelect1(int[] list, int k) {
    selectHelper1(list, 0, list.length - 1, k);

public static void selectHelper1(int[] list, int left, int right, int k) {
    int pivotIndex = partition(list, left, right);
    if (pivotIndex == k) {
    } else if (k < pivotIndex) {
        selectHelper1(list, left, pivotIndex - 1, k);
    } else {
        selectHelper1(list, pivotIndex + 1, right, k);

private static int partition(int[] list, int left, int right) {
    int pivot = left + (right - left) / 2;
    swap(list, right, pivot);
    for (int i = left; i < right; i++) {
        if (list[i] < list[right]) {
            swap(list, i, left);
    swap(list, left, right);
    return left;

quickselect, iteratively

public static int quickSelect2(int[] arr, int k) {
    if (arr == null || arr.length <= k)
        throw new Error();
    int from = 0, to = arr.length - 1;
    // if from == to we reached the kth element
    while (from < to) {
        int r = from, w = to;
        int mid = arr[(r + w) / 2];
        // stop if the reader and writer meets
        while (r < w) {
            if (arr[r] >= mid) { // put the large values at the end
                swap(arr, w, r);
            } else { // the value is smaller than the pivot, skip
        // if we stepped up (r++) we need to step one down
        if (arr[r] > mid)
        // the r pointer is on the end of the first k elements
        if (k <= r) {
            to = r;
        } else {
            from = r + 1;
    return arr[k];