Incrementable Stack - Microsoft Top Interview Questions

Problem Statement :

Implement a stack with the following methods:

IncrementableStack() constructs a new instance of an incrementable stack

append(int val) appends val to the stack

pop() pops and returns the last element in the stack

increment(int inc, inc count) increments the value of bottom count elements by inc

Each method should be done in \mathcal{O}(1)O(1) time. You can assume that for pop, the stack is 
non-empty when it is called.


n ≤ 100,000 where n is the number of calls to append, pop, and increment`.

Example 1


methods = ["constructor", "append", "append", "increment", "pop", "pop"]

arguments = [[], [1], [2], [5, 1], [], []]`


[None, None, None, None, 2, 6]

Solution :


                        Solution in C++ :

class IncrementableStack {
    // {value, increment}
    // all the elements below this element will get affected by the increment value
    vector<pair<int, int>> s;
    IncrementableStack() {

    void append(int val) {
        s.push_back({val, 0});

    int pop() {
        int val = s.back().first;
        int inc = s.back().second;
        int ret = val + inc;
        if (s.size() > 1) s[s.size() - 2].second += inc;
        return ret;

    void increment(int inc, int count) {
        if (s.size() && count) {
            int index = min(count - 1, (int)s.size() - 1);
            s[index].second += inc;

                        Solution in Java :

import java.util.*;

class IncrementableStack {
    // first num is val, second is increment
    LinkedList<int[]> stack = new LinkedList<>();

    public void append(int val) {
        stack.add(new int[] {val, 0});

    public int pop() {
        // Get top of stack
        int[] pair = stack.getLast();
        int val = pair[0], incr = pair[1];

        // Update next pair in stack
        int len = stack.size();
        if (len > 1) {
            stack.get(len - 2)[1] += incr;

        // Remove top of stack

        return val + incr;

    public void increment(int inc, int count) {
        // Increase increment of desired position count
        if (!stack.isEmpty() && count > 0) {
            int index = Math.min(stack.size() - 1, count - 1);
            stack.get(index)[1] += inc;

                        Solution in Python : 
class IncrementableStack:
    def __init__(self):
        self.stack = []
        self.incs = []

    def append(self, val):

    def pop(self):
        inc = self.incs.pop()
        if self.incs:
            self.incs[-1] += inc
        return self.stack.pop() + inc

    def increment(self, inc, count):
        if self.incs and count > 0:
            if count <= len(self.incs):
                self.incs[count - 1] += inc
                self.incs[-1] += inc

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