Watch 10 video solutions for Design a Stack With Increment Operation, a medium level problem involving Array, Stack, Design. This walkthrough by Light Of Truth has 1,282 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Design a stack that supports increment operations on its elements.
Implement the CustomStack class:
CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack.void push(int x) Adds x to the top of the stack if the stack has not reached the maxSize.int pop() Pops and returns the top of the stack or -1 if the stack is empty.void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, increment all the elements in the stack.
Example 1:
Input ["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"] [[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]] Output [null,null,null,2,null,null,null,null,null,103,202,201,-1] Explanation CustomStack stk = new CustomStack(3); // Stack is Empty [] stk.push(1); // stack becomes [1] stk.push(2); // stack becomes [1, 2] stk.pop(); // return 2 --> Return top of the stack 2, stack becomes [1] stk.push(2); // stack becomes [1, 2] stk.push(3); // stack becomes [1, 2, 3] stk.push(4); // stack still [1, 2, 3], Do not add another elements as size is 4 stk.increment(5, 100); // stack becomes [101, 102, 103] stk.increment(2, 100); // stack becomes [201, 202, 103] stk.pop(); // return 103 --> Return top of the stack 103, stack becomes [201, 202] stk.pop(); // return 202 --> Return top of the stack 202, stack becomes [201] stk.pop(); // return 201 --> Return top of the stack 201, stack becomes [] stk.pop(); // return -1 --> Stack is empty return -1.
Constraints:
1 <= maxSize, x, k <= 10000 <= val <= 1001000 calls will be made to each method of increment, push and pop each separately.