Insertion Sort Algorithm

AlgoLab - Sorting & Searching Algorithm Laboratory

• Intuition :

Insertion Sort takes one element at a time and inserts into the correct position in the already sorted part.

• Step — by — Step explanation:

For example, [ 5 , 3 , 4 , 1 ]

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• Take 3 → insert before 5 → [3, 5, 4, 1]
• Take 4 → insert between 3 and 5 → [3, 4, 5, 1]
• Take 1 → insert at beginning → [1, 3, 4, 5]
  Step 1: Start from index 1
  for (int i = 1; i < arr.length; i++)
  We start from i = 1 because the first element 5 is considered already sorted.
  [5 | 3 4 1]
  ↑
  sorted part.
  Step 2: Store current element
  int key = arr[i];
  For i = 1:
  key = 3
  We want to find the correct position for 3.
  Step 3: Take previous index
  int j = i - 1;
  So:
  j = 0
  arr[j] = 5
  Step 4: Compare
  while (j >= 0 && arr[j] > key)
  Check:
  5 > 3 → true
  So we move 5 one position right.
  arr[j + 1] = arr[j];
  Array becomes:
  [5, 5, 4, 1]
  Then:
  j--;
  Now j = -1.
  Step 5: Insert the key
  arr[j + 1] = key;
  Since j = -1:
  arr[0] = 3
  Array:
  [3, 5, 4, 1]
  Step 6: Next element
  Now:
  [3, 5 | 4 1]
  key = 4
  Compare:
  5 > 4 → shift 5
  [3, 5, 5, 1]
  Then insert 4:
  [3, 4, 5, 1]
  Step 7: Last element
  [3, 4, 5 | 1]
  key = 1
  Shift:
  5 → right
  4 → right
  3 → right
  Then insert 1:
  [1, 3, 4, 5]
• Code in java :

import java.util.Scanner;

public class InsertionSort {

    static void insertionSort(int[] arr) {
        for (int i = 1; i < arr.length; i++) {
            int key = arr[i];
            int j = i - 1;

            while (j >= 0 && arr[j] > key) {
                arr[j + 1] = arr[j];
                j--;
            }

            arr[j + 1] = key;
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int[] arr = new int[n];

        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }

        insertionSort(arr);

        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}

• Code in C + +:

#include <iostream>
using namespace std;

void insertionSort(int arr[], int n) {

    for (int i = 1; i < n; i++) {
        int key = arr[i];
        int j = i - 1;

        while (j >= 0 && arr[j] > key) {
            arr[j + 1] = arr[j];
            j--;
        }

        arr[j + 1] = key;
    }
}

int main() {
    int n;
    cin >> n;

    int arr[n];

    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }

    insertionSort(arr, n);

    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }

    return 0;
}

• Time and Space Complexity:

Time: Best O(n) when the array is already sorted, because each element needs only one comparison. Average and worst case are O(n²) because elements may need to be shifted many positions.
Space: O(1) because only key and j are used as extra variables, so no additional array is required

I hope you will understand……best of luck on your journey .