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.
Constraints
n ≤ 100,000 where n is the number of calls to append, pop, and increment`.
Example 1
Input
methods = ["constructor", "append", "append", "increment", "pop", "pop"]
arguments = [[], [1], [2], [5, 1], [], []]`
Output
[None, None, None, None, 2, 6]
Solution :
Solution in C++ :
class IncrementableStack {
public:
// {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;
s.pop_back();
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
stack.removeLast();
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):
self.stack.append(val)
self.incs.append(0)
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
else:
self.incs[-1] += inc
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