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  1. Practice
  2. Algorithms
  3. Sorting
  4. Insertion Sort - Part 2
  5. Discussions

Insertion Sort - Part 2

  • RodneyShag + 3 comments

    Java solution - passes 100% of test cases

    public static void insertionSortPart2(int[] array) {
    for (int i = 1; i < array.length; i++) {
        int j = i;
        int value = array[i];
        while (j >= 1 && array[j-1] > value) {
            array[j] = array[j-1];
            j--;
        }
        array[j] = value;
        printArray(array);
    }
}

    Full solution available in my HackerRank solutions.

    Let me know if you have any questions.

    • adamchenwei + 1 comment

      Hey, converted your answer to JS, but it will only pass first 2 test, thought?

      for (var i = 1; i < array.length; i++) {
  var j = i;
  var value = array[i];
  while (j >= 1 && (array[j-1] > value)) {
    array[j] = array[j-1];
    j--;
  }
  array[j] = value;
  console.log(array.join(' '));
}
      • RodneyShag + 1 comment

        Looks identical. My full solution also prints a newline after each array. My best guess is to make sure the output looks like the expected output in the testcase.

        HackerRank solutions.

        • tawriffic + 1 comment

          It might be sorting the numbers 'alphabetically' instead of 'numerically'....my JS code only passed the first two tests until I realised this. Need the array contents to be integers not string characters. So my array was looking like 2,20,23,26,3,4,5,52, 6 etc

          • RodneyShag + 0 comments

            Nice find!

    • Vasu1819 + 1 comment

      Hi, I've written a similar code but it failed 3/5 test cases reason being error with ArrayOutOfBound exception. Can you please tell me what wrong did I do? Thanks.

      for(int j=1;j<ar.length;j++){
                int temp=ar[j];
                int k = j-1;
                while(ar[k]>temp&&k>=0)
                {
                    ar[k+1]=ar[k];
                    k--;
                }
                ar[k+1]=temp;
                printArray(ar);
            }
      • RodneyShag + 1 comment

        What's the input of the test case that's failing?

        • Vasu1819 + 1 comment

          Test cases 1,2and4. Input of test case 1 being: 9 9 8 6 7 3 5 4 1 2

          • RodneyShag + 2 comments

            Hi. The problem is with this line of your code:

            while(ar[k]>temp&&k>=0)

            It evaluates the expression inside the () from left to right. So it first tries to do ar[k], where k is -1 and it gets an Index Out of Bounds Exception. To fix it, you can change it to

            while(k >= 0 && ar[k] > temp)

            Now, it will check for k >= 0 first. If it's not greater than 0, the next expression (to the right of &&) will not evaluate.

            This is called short circuit evaluation

            HackerRank solutions.

            • Vasu1819 + 0 comments

              Man you are awesome. You made my day. I didn't know about this, thank you. Thank you so much :-)

            • tripathi_shubha1 + 0 comments

              thanku so much !!

    • RA1511003010711 + 1 comment

      is there any need to use "value"..? Can't we simply write array[j-1]>array[i] ..?

      • RodneyShag + 0 comments

        You need to save array[i], since the while loop will be overwriting it.

        HackerRank solutions.