Arrays are one of the most used data structures. You probably have used it a lot already. But, are you aware of the runtimes of push, splice, shift, indexOf, and other operations? In this chapter, we are going deeper into the most common operations and their runtimes.
An array is a collection of things (strings, characters, numbers, objects, etc.). They can be many or zero.
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Tip
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Strings are a collection of characters. Most of the array methods apply to strings as well. |
Arrays look like this:
Arrays are a contiguous collection of elements that can be accessed randomly using an index. This access by index operation takes O(1) time. Let’s take a look at the different functions that we can do with arrays.
const array = [2, 5, 1, 9, 6, 7];
const string = "hello";
console.log(array[2]); // 1
console.log(string[1]); // "e"As you can see, you can access the string’s characters using the same operator as arrays.
You can update arrays in the same way, using the [] operator. However, you can’t modify strings. They are immutable!
const array = [2, 5, 1, 9, 6, 7];
const string = "hello";
array[2] = 117;
console.log(array[2]); // 117
string[1] = "z"; // doesn't change the string.
console.log(string[1]); // "e"|
Warning
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When you try to modify and string, you won’t get an error or anything. It just gets ignored! Your only option is to create a new string with the adjusted value. |
Insertions on an array have different times complexities. O(1): constant time (on average) to append a value at the end of the array. O(n): linear time to insert a value at the beginning or middle.
What if you want to insert a new element at the beginning of the array? You would have to push every item to the right. We can use the following method:
const newArrLength = arr.unshift(element1[, ...[, elementN]]);Here’s an example:
const array = [2, 5, 1];
array.unshift(0); // ↪️ 4
console.log(array); // [ 0, 2, 5, 1 ]
array.unshift(-2, -1); // ↪️ 6
console.log(array); // [ -2, -1, 0, 2, 5, 1 ]As you can see, 2 was at index 0, now was pushed to index 1, and everything else is on a different index. unshift takes O(n) since it affects all the elements of the array.
array.unshiftThe unshift() method adds one or more elements to the beginning of an array and returns its new length.
Runtime: O(n).
Inserting a new element in the middle involves moving part of the array but not all of the items. We can use splice for that:
const arrDeletedItems = arr.splice(start[, deleteCount[, item1[, item2[, ...]]]]);Based on the parameters it takes, you can see that we can add and delete items. Here’s an example of inserting in the middle.
const array = [2, 5, 1, 9, 6, 7];
array.splice(1, 0, 111); // ↪️ [] (1)
// array: [2, 111, 5, 1, 9, 6, 7]-
at position
1, delete0elements and insert111.
The Big O for this operation would be O(n) since, in the worst case, it would move most of the elements to the right.
array.spliceThe splice() method changes an array’s contents by removing existing elements or adding new items. Splice returns an array containing the deleted items.
Runtime: O(n).
For inserting items at the end of the array, we can use: push.
const newArrLength = arr.push([element1[, ...[, elementN]]]);We can push new values to the end of the array like this:
const array = [2, 5, 1, 9, 6, 7];
array.push(4); // ↪️ 7 (1)
// array: [2, 5, 1, 9, 6, 7, 4]-
The
4element would be pushed to the end of the array. Notice thatpushreturns the new length of the array.
Adding to the tail of the array doesn’t change other indexes. E.g., element 2 is still at index 0. So, this is a constant time operation O(1).
array.pushThe push() method adds one or more elements to the end of an array and returns its new length.
Runtime: O(1).
As we saw before, searching by the index is very easy using the [] operator:
const array = [2, 5, 1, 9, 6, 7];
array[4]; // ↪️ 6Searching by index takes constant time - O(1) - to retrieve values out of the array.
Searching by value can be done using indexOf.
const index = arr.indexOf(searchElement[, fromIndex]);If the value is there, we will get the index, otherwise -1.
const array = [2, 5, 1, 9, 6, 7];
console.log(array.indexOf(9)); // ↪️ 3
console.log(array.indexOf(90)); // ↪️ -1Internally, indexOf has to loop through the whole array (worst case) or until we find the first occurrence. Time complexity is O(n).
There are three possible deletion scenarios (similar to insertion): removing at the beginning, middle, or end.
Deleting from the beginning can be done using the splice function and the shift. For simplicity, we will use the latter.
const removedElement = arr.shift();
let arrDeletedItems = arr.splice(start[, deleteCount[, item1[, item2[, ...]]]]);const array = [2, 111, 5, 1, 9, 6, 7];
// Deleting from the beginning of the array.
array.shift(); // ↪️ 2
array.shift(); // ↪️ 111
console.log(array); // [5, 1, 9, 6, 7]
array.splice(0, 1); // ↪️ [ 5 ]
console.log(array); // [ 1, 9, 6, 7 ]As expected, this will change every index on the array, so this takes linear time: O(n).
The shift() method shift all elements to the left. In turn, it removes the first element from an array and returns that removed element. This method changes the length of the array.
Runtime: O(n).
We can use the splice method for deleting an item from the middle of an array.
You can delete multiple items at once:
const array = [0, 1, 2, 3, 4];
// Deleting from the middle
array.splice(2, 3); // ↪️ [ 2, 3, 4 ] (1)
console.log(array); // [0, 1]-
delete 3 elements starting on position 2
Deleting from the middle might cause most of the array elements to move up one position to fill in for the eliminated item. Thus, runtime: O(n).
Removing the last element is very straightforward using pop:
const removedItem = arr.pop();Here’s an example:
const array = [2, 5, 1, 9, 111];
array.pop(); // ↪️111
// array: [2, 5, 1, 9]While deleting the last element, no other item was touched, so that’s an O(1) runtime.
array.popThe pop() method removes the last element from an array and returns that element. This method changes the length of the array.
Runtime: O(1).
To sum up, the time complexity of an array is:
Data Structure |
Searching By |
Inserting at the |
Deleting from |
Space |
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Index/Key |
Value |
beginning |
middle |
end |
beginning |
middle |
end |
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Array |
O(1) |
O(n) |
O(n) |
O(n) |
O(1) |
O(n) |
O(n) |
O(1) |
O(n) |
Operation |
Time Complexity |
Usage |
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O(1) |
Insert element on the right side. |
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O(1) |
Remove the rightmost element. |
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O(1) |
Search for element by index. |
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O(n) |
Search for element by value. |
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O(n) |
Insert element on the left side. |
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O(n) |
Remove leftmost element. |
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O(n) |
Insert and remove from anywhere. |
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O(n) |
Returns a shallow copy of the array. |
Many programming problems involve manipulating arrays. Here are some patterns that can help you improve your problem-solving skills.
Usually, we use one pointer to navigate each element in an array. However, there are times when having two pointers (left/right, low/high) comes in handy. Let’s do some examples.
AR-A) Given a sorted array of integers, find two numbers that add up to a target and return their values.
/**
* Find two numbers that add up target.
* @param arr - The array of integers
* @param target - The target
* @returns {number[]} - array with the values that add up to target.
*/
function twoSum(arr, target) {
// give it a try on your own ...
}twoSum([ -5, -3, 1, 10 ], 7); // [-3, 10] // (10 - 3 = 7)
twoSum([ -5, -3, -1, 1, 2 ], 30); // [] // no 2 numbers add up to 30
twoSum([ -3, -2, -1, 1, 1, 3, 4], -4); // [-3, -1] // (-3 -1 = -4)Solutions:
One naive solution would be to use two pointers in a nested loop:
function twoSum(arr, target) {
for (let i = 0; i < arr.length - 1; i++)
for (let j = i + 1; j < arr.length; j++)
if (arr[i] + arr[j] === target) return [arr[i], arr[j]];
return [];
}The runtime of this solution would be O(n^2). Because of the nested loops. Can we do better? We are not using the fact that the array is SORTED!
We can use two pointers: one pointer starting from the left side and the other from the right side.
Depending on whether the sum is bigger or smaller than the target, we move right or left. If the sum is equal to the target, we return the current left and right pointer’s values.
function twoSum(arr, target) {
let left = 0, right = arr.length -1;
while (left < right) {
const sum = arr[left] + arr[right];
if (sum === target) return [arr[left], arr[right]];
else if (sum > target) right--;
else left++;
}
return [];
}These two pointers have a runtime of O(n).
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Warning
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This technique only works for sorted arrays. If the array was not sorted, you would have to sort it first or choose another approach. |
The sliding window pattern is similar to the two pointers. The difference is that the distance between the left and right pointer is always the same. Also, the numbers don’t need to be sorted. Let’s do an example!
AR-B) Find the max sum of an array of integers, only taking k items from the right and left side sequentially.
Constraints: k won’t exceed the number of elements in the array: 1 ⇐ k ⇐ n.
/**
* Find the max sum of an array of integers,
* only taking `k` items from the right and left side.
*
* @param {number[]} arr - The array of integers
* @param {number} k - The number of elements to sum up.
* @returns {number}
*/
function maxSum(arr, k) {
// Give it a try
};maxSum([1,2,3], 3); // 6 // (1 + 2 + 3 = 6)
maxSum([1,1,1,1,200,1], 3); // 202 // (1 + 200 + 1 = 202)
maxSum([3, 10, 12, 4, 7, 2, 100, 1], 3); // 104 // (3 + 1 + 100 = 104)
maxSum([1,200,1], 1); // 6 // (1 + 2 + 3 = 6)There are multiple ways to solve this problem. Before applying the sliding window, let’s consider this other algorithm:
Backtracking algorithm
Let’s take [3, 10, 12, 4, 7, 2, 100, 1], k = 3 as an example.
-
We have two initial choices: going left with
3or right with1. -
We can take the first element from the left side
3; from there, we can keep going left with10or right1. -
If we go right with
1on the right side, next, we have two options from the right side100or10. -
If we go with
100, then we compute the sum3 + 1 + 100 = 104. -
Repeat with other combinations and keep track of the max sum.
How many combinations can we form? 2^k, since in the worst-case k is equal to n, then we have a runtime of O(2^n)!
We can also visualize all the options as follows. If you add up the numbers from top to bottom, you get the result for all combinations:
graph G {
0 -- 3
0 -- 1
3 -- 10
3 -- a1
10 -- 12
10 -- b1
a1 -- a10
a1 -- 100
1 -- a3
1 -- a100
a3 -- b10
a3 -- b100
a100 -- b3
a100 -- 2
1, a1, b1 [label = 1 color = red]
10, a10, b10 [label = 10 color = red]
3, a3, b3 [label = 3 color = red]
100, a100, b100 [label = 100 color = red]
12 -- res1 [color = gray]
b1 -- res2 [color = gray]
a10 -- res3 [color = gray]
100 -- res4 [color = gray]
b10 -- res5 [color = gray]
b100 -- res6 [color = gray]
b3 -- res7 [color = gray]
2 -- res8 [color = gray]
res1 [label = "Sum: 25", shape=plaintext, fontcolor=gray]
res2 [label = "Sum: 14", shape=plaintext, fontcolor=gray]
res3 [label = "Sum: 14", shape=plaintext, fontcolor=gray]
res4 [label = "Sum: 104", shape=plaintext, fontcolor=gray]
res5 [label = "Sum: 14", shape=plaintext, fontcolor=gray]
res6 [label = "Sum: 104", shape=plaintext, fontcolor=gray]
res7 [label = "Sum: 104", shape=plaintext, fontcolor=gray]
res8 [label = "Sum: 103", shape=plaintext, fontcolor=gray]
}
Notice that many middle branches (in red color) have the same numbers, but in a different order, so their sums oscillate between 104 and 14. That’s why this algorithm is not very optimal for this problem.
Sliding window algorithm
Another approach is using sliding windows. Since the sum always has k elements, we can compute the cumulative sum for the k first elements from the left. Then, we slide the "window" to the right and remove one from the left until we cover all the right items. In the end, we would have all the possible combinations without duplicated work.
Check out the following illustration:
Here’s the implementation:
function maxSum(arr, k) {
let left = k - 1;
let right = arr.length -1;
let sum = 0;
for (let i = 0; i < k; i++) sum += arr[i];
let max = sum;
for (let i = 0; i < k; i++) {
sum += arr[right--] - arr[left--];
max = Math.max(max, sum);
}
return max;
};The difference between the two pointers pattern and the sliding windows, it’s that we move both pointers at the same time to keep the length of the window the same.
The runtime for this code is: k. As we move the window k times. Since k ⇐ n, the final runtime is O(n).
AR-1) Given an array of integers, find the maximum sum of consecutive elements (subarray).
Examples:
maxSubArray([1, -3, 10, -5]); // 10 (taking only 10)
maxSubArray([-3, 4,-1, 2, 1, -5]); // 6 (sum [4,-1, 2, 1])
maxSubArray([-2, 1, -3, 4, -1, 3, 1]); // 7 (sum [4,-1, 3, 1])Common in interviews at: FAANG, Microsoft
link:../../interview-questions/max-subarray.js[role=include]
// write you code here
}Solution: [array-q-max-subarray]
AR-2) You have an array of integers. Each value represents the closing value of the stock on that day. You have only one chance to buy and then sell. What’s the maximum profit you can obtain? (Note: you have to buy first and then sell)
Examples:
maxProfit([1, 2, 3]) // 2 (buying at 1 and selling at 3)
maxProfit([3, 2, 1]) // 2 (no buys)
maxProfit([5, 10, 5, 10]) // 5 (buying at 5 and selling at 10)Common in interviews at: Amazon, Facebook, Bloomberg
link:../../interview-questions/buy-sell-stock.js[role=include]
// write you code here
}Solution: [array-q-buy-sell-stock]