2011. Final Value of Variable After Performing Operations
Problem Description
The problem presents a simple programming language with a single variable X and four operations that can modify the value of X. These operations are incrementing X by 1 (++X or X++) and decrementing X by 1 (--X or X--). Initially, X is 0. The task is to determine the final value of X after executing a sequence of these operations provided in an array of strings.
Intuition
The solution approach is quite straightforward due to the problem's simplicity. Since there are only two types of operations that affect X - either incrementing or decrementing by 1, we can iterate through the array of operations and simply count how many times each operation occurs. Since it's given that the operations are strings containing either ++ or --, we can inspect the second character of each operation string to determine whether we're dealing with an increment or a decrement operation. This is a valid approach because the second character explicitly determines operation type: if it's a +, we increment; if it's a -, we decrement. We do not need to differentiate between pre- or post-increment/decrement operations for this particular problem since the effect on the final value of X is the same.
By looping through the array once and summing the value of 1 for increments and -1 for decrements, we effectively calculate the final value of X after all operations have been applied. The sum of 1s and -1s then gives us the direct final result without needing to keep track of the value of X after each individual operation.
What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?
1def fun(arr: List[int]) -> List[int]:
2 import heapq
3 heapq.heapify(arr)
4 res = []
5 for i in range(3):
6 res.append(heapq.heappop(arr))
7 return res
81public static int[] fun(int[] arr) {
2 int[] res = new int[3];
3 PriorityQueue<Integer> heap = new PriorityQueue<>();
4 for (int i = 0; i < arr.length; i++) {
5 heap.add(arr[i]);
6 }
7 for (int i = 0; i < 3; i++) {
8 res[i] = heap.poll();
9 }
10 return res;
11}
121class HeapItem {
2 constructor(item, priority = item) {
3 this.item = item;
4 this.priority = priority;
5 }
6}
7
8class MinHeap {
9 constructor() {
10 this.heap = [];
11 }
12
13 push(node) {
14 // insert the new node at the end of the heap array
15 this.heap.push(node);
16 // find the correct position for the new node
17 this.bubble_up();
18 }
19
20 bubble_up() {
21 let index = this.heap.length - 1;
22
23 while (index > 0) {
24 const element = this.heap[index];
25 const parentIndex = Math.floor((index - 1) / 2);
26 const parent = this.heap[parentIndex];
27
28 if (parent.priority <= element.priority) break;
29 // if the parent is bigger than the child then swap the parent and child
30 this.heap[index] = parent;
31 this.heap[parentIndex] = element;
32 index = parentIndex;
33 }
34 }
35
36 pop() {
37 const min = this.heap[0];
38 this.heap[0] = this.heap[this.size() - 1];
39 this.heap.pop();
40 this.bubble_down();
41 return min;
42 }
43
44 bubble_down() {
45 let index = 0;
46 let min = index;
47 const n = this.heap.length;
48
49 while (index < n) {
50 const left = 2 * index + 1;
51 const right = left + 1;
52
53 if (left < n && this.heap[left].priority < this.heap[min].priority) {
54 min = left;
55 }
56 if (right < n && this.heap[right].priority < this.heap[min].priority) {
57 min = right;
58 }
59 if (min === index) break;
60 [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61 index = min;
62 }
63 }
64
65 peek() {
66 return this.heap[0];
67 }
68
69 size() {
70 return this.heap.length;
71 }
72}
73
74function fun(arr) {
75 const heap = new MinHeap();
76 for (const x of arr) {
77 heap.push(new HeapItem(x));
78 }
79 const res = [];
80 for (let i = 0; i < 3; i++) {
81 res.push(heap.pop().item);
82 }
83 return res;
84}
85Solution Approach
The implementation of the solution utilizes a one-pass loop through the list of operation strings. The Python code takes advantage of list comprehension—a compact syntax for processing lists—and the sum function to aggregate the results.
Here is a step-by-step explanation of the solution approach:
- List Comprehension: The
for s in operationspart of the list comprehension iterates over each stringsin theoperationslist. - Condition Check: For each string
s,s[1] == '+'checks whether the second character is a'+'. The comparison yields a boolean value (Truefor increment,Falsefor decrement). - Ternary Operation: Based on the result of the condition check,
1 if s[1] == '+' else -1evaluates to1if the character is a'+', meaning the operation is an increment. If the character is not a'+'(thus a'-'since those are the only options), the result is-1for a decrement operation. - Summation: The
sum(...)function takes all these1s and-1s and sums them up to get the total net effect onX.
The solution does not make use of any additional data structures, resulting in a space complexity of O(1). The time complexity is O(n) where n is the number of operations, since it only involves iterating through the list of operations once.
No complex algorithms or design patterns are necessary because we simply transform each operation into its numeric effect (+1 or -1) and then summarize these effects to get the final result.
Given an array of 1,000,000 integers that is almost sorted, except for 2 pairs of integers. Which algorithm is fastest for sorting the array?
Example Walkthrough
Let's consider a small example to illustrate the solution approach. Suppose the array of operations is as follows:
1operations = ["++X", "X++", "--X", "X--", "X++"]
Following the solution approach step by step:
-
Initialize a sum variable to 0 which will hold the final value of
X. -
Iterate over each operation in the list:
-
The first operation is "++X" which has a '+' as its second character. According to the approach,
Xshould be incremented by 1. -
The second operation is "X++" which also has a '+' as its second character. Again,
Xshould be incremented by 1. -
The third operation is "--X" which has a '-' as its second character. This time,
Xshould be decremented by 1. -
The fourth operation is "X--" which again has a '-' as its second character. We decrement
Xby 1 once more. -
The fifth operation is "X++" with a '+' as the second character, and
Xis incremented by 1.
-
-
Apply the operations to the sum variable:
- After the first operation, sum = 0 + 1 = 1
- After the second operation, sum = 1 + 1 = 2
- Following the third operation, sum = 2 - 1 = 1
- After the fourth operation, sum = 1 - 1 = 0
- After the fifth operation, sum = 0 + 1 = 1
-
We have now completed iterating through the operations, and the final sum value of 1 represents the final value of
X.
Using the list comprehension and ternary operation, the Python code using this solution approach would look like this:
1final_value_of_X = sum(1 if s[1] == '+' else -1 for s in operations)
And for our example array, executing this code will result in final_value_of_X being 1, which matches our manual calculation.
Solution Implementation
1from typing import List # Need to import List for type hints
2
3class Solution:
4 def finalValueAfterOperations(self, operations: List[str]) -> int:
5 """
6 This function computes the final value after performing all the operations in the given list.
7
8 Parameters:
9 operations (List[str]): A list of string operations, each element is one of "++X", "X++", "--X", "X--".
10
11 Returns:
12 int: The final integer value after all operations are performed.
13 """
14
15 # Initialize the final value to 0
16 final_value = 0
17
18 # Loop through the list of operations
19 for operation in operations:
20 # Increment final_value for operations that have a '+' in the middle (i.e. '++X' or 'X++')
21 if operation[1] == '+':
22 final_value += 1
23 # Decrement final_value for operations that have a '-' in the middle (i.e. '--X' or 'X--')
24 else:
25 final_value -= 1
26
27 # Return the calculated final value
28 return final_value
29
30# Example usage:
31# solution = Solution()
32# result = solution.finalValueAfterOperations(["--X", "X++", "X++"])
33# print(result) # Output would be 1, since the final value is incremented twice and decremented once.
341class Solution {
2
3 // This method computes the final value after performing all the operations in the array.
4 public int finalValueAfterOperations(String[] operations) {
5 // Initialize the result to 0.
6 int result = 0;
7
8 // Loop through each operation string in the operations array.
9 for (String operation : operations) {
10 // Check the second character of the operation string to determine if it's an increment or decrement.
11 // The second character will be '+' for increment and '-' for decrement operations.
12 if (operation.charAt(1) == '+') {
13 // If we have an increment operation, increase the result by 1.
14 result++;
15 } else {
16 // If we have a decrement operation, decrease the result by 1.
17 result--;
18 }
19 }
20
21 // Return the final result after performing all operations.
22 return result;
23 }
24}
251#include <vector>
2#include <string>
3
4class Solution {
5public:
6 // Function to calculate the final value after performing all operations.
7 int finalValueAfterOperations(std::vector<std::string>& operations) {
8 int finalValue = 0; // Initialize the final value to 0.
9
10 // Loop over each operation string in the vector.
11 for (const auto& operation : operations) {
12 // Check the second character of the string to determine operation type.
13 if (operation[1] == '+') {
14 finalValue++; // If it's a plus, increment the value.
15 } else {
16 finalValue--; // If it's a minus, decrement the value.
17 }
18 }
19
20 // Return the result after performing all operations.
21 return finalValue;
22 }
23};
241// Defines a function that calculates the final value after processing all operations
2// @param operations - An array of strings representing operations, each being either "++X", "X++", "--X", or "X--"
3// @returns The final value after performing all operations
4function finalValueAfterOperations(operations: string[]): number {
5 // Reduce the operations array to a single value that represents the final value
6 // @param accumulator - The running total of the operations processed
7 // @param currentValue - The current operation string being processed
8 // @returns The updated running total after processing the current operation
9 return operations.reduce((accumulator, currentValue) => {
10 // Check if the second character of the operation string is a '+' or '-'
11 // Increment or decrement the accumulator accordingly
12 return accumulator + (currentValue[1] === '+' ? 1 : -1);
13 }, 0); // Initialize the accumulator to 0, since the starting value is 0
14}
15
16// Example usage of the function:
17// const result = finalValueAfterOperations(["--X", "X++", "X++"]);
18// console.log(result); // Output will be 1
19Given an array of 1,000,000 integers that is almost sorted, except for 2 pairs of integers. Which algorithm is fastest for sorting the array?
Time and Space Complexity
Time Complexity
The time complexity of the provided code is O(n), where n is the number of operations in the input list. This is because the code iterates through each element in the operations list exactly once to determine if the operation is an increment or decrement.
Space Complexity
The space complexity of the provided code is O(1). The only extra space used is for the accumulator in the sum operation which keeps track of the final result, regardless of the number of operations.
Learn more about how to find time and space complexity quickly using problem constraints.
Which data structure is used to implement priority queue?
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