We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Sorting
  4. Insertion Sort - Part 1
  5. Discussions

Insertion Sort - Part 1

Sort by

recency

|

896 Discussions

|

  • M_elmasry
     + 0 comments

    this my python solution :.

    def insertionSort1(n, arr): m = arr[n - 1]

    for j in range(n - 1, 0, -1):
    if arr[j - 1] > m:
        arr[j] = arr[j - 1]
        print(' '.join(str(x) for x in arr))
    else:
        arr[j] = m
        print(' '.join(str(x) for x in arr))
        return  # Exit after inserting

# Insert at the beginning if m is the smallest
arr[0] = m
print(' '.join(str(x) for x in arr))

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanatoin here : https://youtu.be/xFqneyHR96g

    void printArr(vector<int> arr){
    for(int i = 0; i < arr.size(); i++){cout << arr[i] << " ";}
    cout << endl;
}
void insertionSort1(int n, vector<int> arr) {
    int last = arr[arr.size()-1];
    for(int i = arr.size() - 1; i >= 0; i--){
        if(last < arr[i-1]) {
            arr[i] = arr[i-1];
            printArr(arr);
        }
        else{
            arr[i] = last;
            printArr(arr);
            break;
        }
    }
}

  • robertbielicki
     + 0 comments

    RUST: 

    fn insertionSort1(n: i32, arr: &[i32]) {
    let mut result = arr.to_vec();
    let number = result.pop().unwrap();
    result.insert((n-1) as usize, *result.last().unwrap());

    for idx in 1..=n {
        if idx < n && result[(n - idx -1) as usize] > number{
            result[(n - idx) as usize] = result[((n - idx - 1)) as usize];
        } else {
            result[(n-idx) as usize] = number;
            let strin = result.iter().map(|&x| x.to_string()).collect::<Vec<String>>().join(" ");
            println!("{}", strin);
            break;  
        }
        let strin = result.iter().map(|&x| x.to_string()).collect::<Vec<String>>().join(" ");
        println!("{}", strin);
    }
}

  • patrickgao2020
     + 0 comments

    Here's my Python solution.

    def insertionSort1(n, arr):
    insert = arr[-1]
    index = list(reversed([i for i in range(len(arr) - 1)]))
    for i in index:
        if insert <= arr[i]:
            arr[i + 1] = arr[i]
            print(" ".join([str(i) for i in arr]))
        else:
            arr[i + 1] = insert
            print(" ".join([str(i) for i in arr]))
            return
    arr[0] = insert
    print(" ".join([str(i) for i in arr]))

  • mrsushilpatel201
     + 1 comment

    Python3: Time Complexity: O(n) (Extra) Space Complexity: O(1)

    def printArray(arr):
    for x in arr:
        print(x, end=' ')
    print()

def insertionSort1(n, arr):
    i = n - 2
    lastElement = arr[n - 1]
    while(i >= 0 and arr[i] > lastElement):
        arr[i + 1] = arr[i]
        printArray(arr)
        i = i - 1
    arr[i + 1] = lastElement
    printArray(arr)