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Commit db9fabc

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stack-2: q & a
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‎book/D-interview-questions-solutions.asc

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@@ -198,7 +198,7 @@ If the index overflows, it moves to the next node and reset the index to zero.
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[#stack-q-valid-parentheses]
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include::content/part02/stack.asc[tag=stack-q-valid-parentheses]
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We need to validate that brackets are properly opened and closed, following these rules:
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.We need to validate that brackets are properly opened and closed, following these rules:
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- An opened bracket must be close by the same type.
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- Open brackets mush be closed in the corrent order.
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- Space: `O(n)`. We use an auxiliary stack.
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// [#linkedlist-q-TAG]
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// include::content/part02/linked-list.asc[tag=linkedlist-q-TAG]
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[#stack-q-daily-temperatures]
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include::content/part02/stack.asc[tag=stack-q-daily-temperatures]
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The first solution that might come to mind it's using two for loops. For each element we count how many elements ahead there's a warmer temperature.
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[source, javascript]
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----
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include::interview-questions/daily-temperatures.js[tag=dailyTemperaturesBrute1]
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----
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This solution is a `O(n^2)`. Can we do better? We can!
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Here's an idea: start backwards so we know when there's a warmer temperature before hand. The last element is always 0 (because there's no more temperatures ahead of it). We can place each element's index that we visit on a stack. If the current temperature is bigger than the on the top of the stack, we remove it until a bigger one remains or the stack is empty. If the stack has a value, we calculate the number of days ahead.
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*Algorithm*:
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- Traverse the daily temperatures backwards
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- Push each temperature to an stack.
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- While the current temperature is larger than the one at the top of the stack, remove it.
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- If the stack is empty, then there's no warmer weather ahead.
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- If the stack has an element, calculate the index delta.
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*Implementation*:
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[source, javascript]
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----
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include::interview-questions/daily-temperatures.js[tag=description]
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include::interview-questions/daily-temperatures.js[tag=solution]
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----
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The stack contains the indexes rather than the temperatures.
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*Complexity Analysis*:
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- Time: `O(n)`. We visit each element on the array once.
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- Space: `O(1)`. The worst case scenario is ascending order without duplicates. The stack will hold at most 70 items (100 - 30). If we didn't have the range restriction then space complexity would be `O(n)`.
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// [#linkedlist-q-FILENAME]
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// include::content/part02/linked-list.asc[tag=linkedlist-q-FILENAME]
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// RESTATE REQUIREMENTS AND DESCRIPTIONS
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// [source, javascript]
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// ----
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// include::interview-questions/TAG.js[tag=description]
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// include::interview-questions/TAG.js[tag=solution]
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// include::interview-questions/FILENAME.js[tag=description]
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// include::interview-questions/FILENAME.js[tag=solution]
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// ----
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// IMPLEMENTATION NOTES

‎book/content/part02/stack.asc

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==== Interview Questions
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(((Interview Questions, Arrays)))
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// tag::stack-q-valid-parentheses[]
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===== Validate Parentheses / Braces / Brackets
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----
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_Solution: <<stack-q-valid-parentheses>>_
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// tag::stack-q-daily-temperatures[]
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===== Daily Temperaturs
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*ST-2*) _Given an array of integers from 30 to 100 (daily temperatures), return another array that for each day in the input, tells you how many days you would have to wait until a warmer temperature. If no warmer temparature is possible then return `0` for that element._
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// end::stack-q-daily-temperatures[]
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[source, javascript]
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----
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include::../../interview-questions/daily-temperatures.js[tag=description]
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// write you code here
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}
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----
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_Solution: <<stack-q-daily-temperatures>>_
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// tag::description[]
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/**
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* Given an array with daily temperatures (30 °C to 100 °C),
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* return an array with the days count until a warmer temperature
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* for each elem from the input.
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*
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* @examples
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* dailyTemperatures([30, 28, 50, 40, 30]); // [2, 1, 0, 0, 0]
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* dailyTemperatures([73, 69, 72, 76, 73]); // [3, 1, 1, 0, 0]
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*
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* @param {number[]} t - Daily temperatures
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*/
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function dailyTemperatures(t) {
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// end::description[]
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// tag::solution[]
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const last = (arr) => arr[arr.length - 1];
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const stack = [];
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const ans = [];
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for (let i = t.length - 1; i >= 0; i--) {
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while (stack.length && t[i] >= t[last(stack)]) stack.pop();
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ans[i] = stack.length ? last(stack) - i : 0;
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stack.push(i);
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}
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return ans;
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}
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// end::solution[]
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// tag::dailyTemperaturesBrute1[]
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function dailyTemperaturesBrute1(t) {
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const ans = [];
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for (let i = 0; i < t.length; i++) {
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ans[i] = 0;
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for (let j = i + 1; j < t.length; j++) {
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if (t[j] > t[i]) {
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ans[i] = j - i;
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break;
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}
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}
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}
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return ans;
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}
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// end::dailyTemperaturesBrute1[]
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module.exports = { dailyTemperatures, dailyTemperaturesBrute1 };
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const { dailyTemperatures } = require('./daily-temperatures');
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describe('Stack: Daily Temperatures', () => {
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it('should work', () => {
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expect(dailyTemperatures([30, 28, 50, 40, 30])).toEqual([2, 1, 0, 0, 0]);
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});
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it('should work', () => {
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expect(dailyTemperatures([73, 74, 75, 71, 69, 72, 76, 73])).toEqual([1, 1, 4, 2, 1, 1, 0, 0]);
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});
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it('should work', () => {
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expect(dailyTemperatures([89, 62, 70, 58, 47, 47, 46, 76, 100, 70])).toEqual([8, 1, 5, 4, 3, 2, 1, 1, 0, 0]);
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});
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it('should work with large data', () => {
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const input = [64, 40, 49, 73, 72, 35, 68, 83, 35, 73, 84, 88, 96, 43, 74, 63, 41, 95, 48, 46, 89, 72, 34, 85, 72, 59, 87, 49, 30, 32, 47, 34, 74, 58, 31, 75, 73, 88, 64, 92, 83, 64, 100, 99, 81, 41, 48, 83, 96, 92, 82, 32, 35, 68, 68, 92, 73, 92, 52, 33, 44, 38, 47, 88, 71, 50, 57, 95, 33, 65, 94, 44, 47, 79, 41, 74, 50, 67, 97, 31, 68, 50, 37, 70, 77, 55, 48, 30, 77, 100, 31, 100, 69, 60, 47, 95, 68, 47, 33, 64];
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const output = [3, 1, 1, 4, 3, 1, 1, 3, 1, 1, 1, 1, 30, 1, 3, 2, 1, 25, 2, 1, 19, 2, 1, 3, 2, 1, 11, 5, 1, 1, 2, 1, 3, 2, 1, 2, 1, 2, 1, 3, 2, 1, 0, 46, 3, 1, 1, 1, 30, 18, 5, 1, 1, 2, 1, 12, 1, 10, 5, 1, 2, 1, 1, 4, 3, 1, 1, 11, 1, 1, 8, 1, 1, 5, 1, 3, 1, 1, 11, 1, 3, 2, 1, 1, 5, 3, 2, 1, 1, 0, 1, 0, 3, 2, 1, 0, 0, 2, 1, 0];
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expect(dailyTemperatures(input)).toEqual(output);
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});
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});

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