Delete Integers In Ascending Order - Google Top Interview Questions

Problem Statement :

You are given a list of unique integers nums and want to delete each number in ascending order.

Return the indices of numbers in order of their deletion.


n ≤ 100,000 where n is the length of nums

Example 1


nums = [3, 5, 1, 4, 2]


[2, 3, 0, 1, 0]

We delete the numbers in this order:

Delete 1 at index 2 and now we have [3, 5, 4, 2]

Delete 2 at index 3 and now we have [3, 5, 4]

Delete 3 at index 0 and now we have [5, 4]

Delete 4 at index 1 and now we have [5]

Delete 5 at index 0 and now we have []

Solution :


                        Solution in C++ :

const int SZ = 1 << 17;
void upd(vector<int>& tree, int idx, int val) {
    idx += SZ;
    while (idx) {
        tree[idx] += val;
        idx /= 2;
int qry(vector<int>& tree, int lhs, int rhs) {
    lhs += SZ;
    rhs += SZ;
    int ret = 0;
    while (lhs <= rhs) {
        if (lhs % 2) ret += tree[lhs++];
        if (rhs % 2 == 0) ret += tree[rhs--];
        lhs /= 2;
        rhs /= 2;
    return ret;
vector<int> solve(vector<int>& nums) {
    vector<int> tree(2 * SZ);
    int n = nums.size();
    vector<int> ret;
    vector<pair<int, int>> pairs;
    for (int i = 0; i < n; i++) pairs.emplace_back(nums[i], i);
    sort(pairs.begin(), pairs.end());
    for (auto p : pairs) {
        ret.push_back(p.second - qry(tree, 0, p.second - 1));
        upd(tree, p.second, 1);
    return ret;

                        Solution in Java :

import java.util.*;

class Solution {
    class SegTree {
        int leftMost, rightMost, sum;
        SegTree left, right;
        int[] a;

        public SegTree(int[] a, int leftMost, int rightMost) {
            this.a = a;
            this.leftMost = leftMost;
            this.rightMost = rightMost;
            int mid = (rightMost + leftMost) / 2;
            if (leftMost == rightMost) {
                sum = a[leftMost];
            } else {
                left = new SegTree(a, leftMost, mid);
                right = new SegTree(a, mid + 1, rightMost);

        void recalc() {
            if (leftMost == rightMost)
            sum = right.sum + left.sum;

        void pointUpdate(int index, int newVal) {
            if (leftMost == rightMost) {
                sum = newVal;
            if (index <= left.rightMost)
                left.pointUpdate(index, newVal);
                right.pointUpdate(index, newVal);

        int rangeSum(int l, int r) {
            // three cases:
            // 1. no intersect
            if (r < leftMost || rightMost < l)
                return 0;
            // 2. partially contained
            if (l <= leftMost && r >= rightMost)
                return sum;
            // 3. don't know
            return left.rangeSum(l, r) + right.rangeSum(l, r);

    int[] solve(int[] nums) {
        List<int[]> ls = new ArrayList<>();
        int[] res = new int[nums.length];
        if (nums.length == 0)
            return res;
        for (int i = 0; i < nums.length; i++) {
            ls.add(new int[] {nums[i], i});

        Collections.sort(ls, (a, b) -> a[0] - b[0]);

        SegTree segTree = new SegTree(new int[nums.length], 0, nums.length - 1);
        for (int i = 0; i < nums.length; i++) {
            int[] n = ls.get(i);
            int index = n[1];
            int rangeSum = segTree.rangeSum(0, index);
            res[i] = index - rangeSum;
            segTree.pointUpdate(index, 1);
        return res;

                        Solution in Python : 
class Solution:
    def solve(self, nums):
        # indices holds pairs (index, number)
        indices = list(enumerate(nums))
        indices.sort(key=lambda x: -x[1])
        sl = SortedList()
        result = []
        for idx, _ in indices:
            t = sl.bisect_left(idx)

        return result[::-1]

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