Selection Sort Algorithm in DSA Javascript - Time & Space Complexity

Time Complexity: Selection Sort Algorithm

O(n²)

Understanding Time Complexity

We want to know how the time taken by selection sort changes as the list gets bigger.

How does the number of steps grow when sorting more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function selectionSort(arr) {
  for (let i = 0; i < arr.length - 1; i++) {
    let minIndex = i;
    for (let j = i + 1; j < arr.length; j++) {
      if (arr[j] < arr[minIndex]) {
        minIndex = j;
      }
    }
    [arr[i], arr[minIndex]] = [arr[minIndex], arr[i]];
  }
  return arr;
}

This code sorts an array by repeatedly finding the smallest item and moving it to the front.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The inner loop compares elements to find the smallest item.
• How many times: The outer loop runs once for each element except the last, and the inner loop runs fewer times each pass but still many times overall.

How Execution Grows With Input

As the list grows, the number of comparisons grows much faster because each item is compared with many others.

Input Size (n)	Approx. Operations
10	About 45 comparisons
100	About 4,950 comparisons
1000	About 499,500 comparisons

Pattern observation: The operations grow roughly by the square of the input size.

Final Time Complexity

Time Complexity: O(n²)

This means if you double the number of items, the work roughly quadruples.

Common Mistake

[X] Wrong: "Selection sort gets faster if the list is already sorted."

[OK] Correct: Selection sort always checks all pairs, so it takes the same time no matter the order.

Interview Connect

Understanding selection sort's time helps you compare it with faster methods and shows you can analyze simple algorithms clearly.

Self-Check

"What if we stopped searching for the minimum after finding the first smaller element? How would the time complexity change?"

Practice

1. What is the main idea behind the Selection Sort algorithm?

easy

A. Divide the array into halves and merge sorted halves

B. Repeatedly find the smallest element and move it to the front

C. Swap adjacent elements if they are in wrong order repeatedly

D. Use a pivot to partition the array into smaller parts

Solution

1. Step 1: Understand Selection Sort process

   Selection Sort works by finding the smallest element in the unsorted part and swapping it with the first unsorted element.

2. Step 2: Compare with other sorting methods

   Other options describe Merge Sort, Bubble Sort, and Quick Sort, which are different algorithms.

3. Final Answer:

   Repeatedly find the smallest element and move it to the front -> Option B

4. Quick Check:

   Selection Sort = Find smallest and swap [OK]

Hint: Selection Sort always picks the smallest next element [OK]

Common Mistakes:

• Confusing Selection Sort with Bubble Sort
• Thinking it uses divide and conquer
• Assuming it uses a pivot element

2. Which of the following is the correct syntax to swap two elements arr[i] and arr[minIndex] in JavaScript during Selection Sort?

easy

A. temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp;

B. arr[i], arr[minIndex] = arr[minIndex], arr[i];

C. [arr[i], arr[minIndex]] = [arr[minIndex], arr[i]];

D. swap(arr[i], arr[minIndex]);

Solution

1. Step 1: Identify JavaScript swap syntax

   JavaScript supports swapping using destructuring assignment: [a, b] = [b, a].

2. Step 2: Check each option

   temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp; is valid but missing let or var for temp and is verbose. arr[i], arr[minIndex] = arr[minIndex], arr[i]; is invalid syntax in JavaScript. swap(arr[i], arr[minIndex]); is a function call but no such built-in function exists.

3. Final Answer:

   [arr[i], arr[minIndex]] = [arr[minIndex], arr[i]]; -> Option C

4. Quick Check:

   Destructuring swap = [arr[i], arr[minIndex]] = [arr[minIndex], arr[i]]; [OK]

Hint: Use destructuring assignment for swapping in JavaScript [OK]

Common Mistakes:

• Using Python-style tuple swap syntax
• Calling undefined swap function
• Forgetting to declare temporary variable

3. What will be the output array after running Selection Sort on [29, 10, 14, 37, 13]?

medium

A. [37, 29, 14, 13, 10]

B. [13, 10, 14, 29, 37]

C. [29, 10, 14, 37, 13]

D. [10, 13, 14, 29, 37]

Solution

1. Step 1: Perform Selection Sort step-by-step

   Initial array: [29, 10, 14, 37, 13]
   Find smallest from index 0 to end: 10 at index 1, swap with 29.
   Array: [10, 29, 14, 37, 13]
   Find smallest from index 1 to end: 13 at index 4, swap with 29.
   Array: [10, 13, 14, 37, 29]
   Find smallest from index 2 to end: 14 at index 2, no swap.
   Find smallest from index 3 to end: 29 at index 4, swap with 37.
   Array: [10, 13, 14, 29, 37]

2. Step 2: Confirm final sorted array

   The array is sorted in ascending order: [10, 13, 14, 29, 37].

3. Final Answer:

   [10, 13, 14, 29, 37] -> Option D

4. Quick Check:

   Sorted array = [10, 13, 14, 29, 37] [OK]

Hint: Selection Sort places smallest element at front each pass [OK]

Common Mistakes:

• Not swapping correctly after finding minimum
• Confusing index positions during swaps
• Assuming array is unchanged

4. Identify the error in this Selection Sort code snippet:

function selectionSort(arr) {
  for (let i = 0; i < arr.length; i++) {
    let minIndex = i;
    for (let j = i + 1; j < arr.length; j++) {
      if (arr[j] < arr[minIndex]) {
        minIndex = j;
      }
    }
    arr[i] = arr[minIndex];
    arr[minIndex] = arr[i];
  }
  return arr;
}

medium

A. Swapping lines overwrite values causing incorrect swaps

B. Loop ranges are incorrect, should be j < arr.length - 1

C. minIndex should be initialized outside the outer loop

D. Function does not return the sorted array

Solution

1. Step 1: Analyze the swap logic

   The swap is done by assigning arr[i] = arr[minIndex] first, then arr[minIndex] = arr[i]. This overwrites arr[i] before assigning to arr[minIndex], causing both to have the same value.

2. Step 2: Correct swap method

   Use a temporary variable or destructuring assignment to swap without overwriting:
   let temp = arr[i]; arr[i] = arr[minIndex]; arr[minIndex] = temp;

3. Final Answer:

   Swapping lines overwrite values causing incorrect swaps -> Option A

4. Quick Check:

   Incorrect swap overwrites values = Swapping lines overwrite values causing incorrect swaps [OK]

Hint: Use temp variable or destructuring to swap safely [OK]

Common Mistakes:

• Swapping without temp variable
• Incorrect loop boundaries
• Not returning the array

5. Given an array with duplicate values [4, 2, 4, 3, 2], how many swaps will Selection Sort perform to fully sort it?

hard

A. 3

B. 5

C. 4

D. 2

Solution

1. Step 1: Trace Selection Sort swaps on the array

   Initial array: [4, 2, 4, 3, 2]
   Pass 1: smallest from index 0-4 is 2 at index 1, swap with 4 at index 0.
   Array: [2, 4, 4, 3, 2] (1 swap)
   Pass 2: smallest from index 1-4 is 2 at index 4, swap with 4 at index 1.
   Array: [2, 2, 4, 3, 4] (2 swaps)
   Pass 3: smallest from index 2-4 is 3 at index 3, swap with 4 at index 2.
   Array: [2, 2, 3, 4, 4] (3 swaps)
   Pass 4: smallest from index 3-4 is 4 at index 3, no swap.
   Pass 5: last element, no swap.

2. Step 2: Count total swaps

   Total swaps performed = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Count swaps during passes = 3 [OK]

Hint: Count swaps only when minimum index differs from current index [OK]

Common Mistakes:

• Counting all comparisons as swaps
• Ignoring swaps when minIndex equals current index
• Assuming duplicates reduce swaps to zero