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  1. Practice
  2. Algorithms
  3. Implementation
  4. Circular Array Rotation
  5. Discussions

Circular Array Rotation

  • murnun + 2 comments

    Nice one liner using the built in functions! Unfortunately, this still fails when k is higher than n (test case #4). You seem to be getting around it by overwriting k with: 

    var k = details[1]%details[0];

    I'm curious to see if there are other, more algorithmic solutions to this in Javascript.

    • matt_heisig + 1 comment

      This works in Javascript for all test cases:

      function main() {
    var n_temp = readLine().split(' ');
    var n = parseInt(n_temp[0]);
    var k = parseInt(n_temp[1]);
    var q = parseInt(n_temp[2]);
  
    a = readLine().split(' ');
    a = a.map(Number);
  
    for (var i=0; i<q; i++) {
      var m = readLine();
      console.log(a.slice(m-k%n)[0]);
    }
}
      • mehul_sachdeva7 + 1 comment

        when i ran my code only sample test cleared and rest failed

        function circularArrayRotation(a, k, queries) { 

    var res = [];
    var resind = 0;

    while (k > 0) {
        rotateArr();
        k--;
    }

    function rotateArr() {
        var i = 0;
        var lastel = a[a.length - 1];
        var temp1 = a[0];
        var temp2;

        while (i < a.length-1) {
            temp2 = a[i + 1];
            a[i + 1] = temp1;
            temp1 = temp2;
            i++;
        }

        a[0] = lastel;
    }

    for(var j in queries) {
        res[resind] = a[j];
        resind++;
    }

    return res;
		}

            I think some issue is in how i am returning the values, please help identify the error
        • mehul_sachdeva7 + 2 comments

          found out what the problem was, new JS code:

          function circularArrayRotation(a, k, queries) {

    var rarr = [];
    var aind = [];
    var res = [];

    for (var i = 0; i < a.length; i++){
        if ((i + k) < a.length) {
            aind[i] = i + k;
        } else {
            aind[i] = (i + k) % a.length;
        }
    }

    for (var j = 0; j < a.length; j++) {
        rarr[aind[j]] = a[j];
    }


    for (var k = 0; k < queries.length; k++){
        res[k] = rarr[queries[k]];
    }

    return (res);
}
          • tashk85 + 0 comments

            Hi, could you please explain what is happening here? Thanks

          • NaveenKashyap + 0 comments

            please explain what is going on in the code

    • gary_l_hewitt + 1 comment

      More algorithmic? - I don't see any reason to mutate the arrays at all, if all you need is the correct indexes from the rotated input. This just plucks the relevant items from the (rotated) indexes in the query:

      function circularArrayRotation(a, k, queries) {
    let offset = a.length - (k % a.length);
    return queries.map( e => a[(e+offset)%a.length]);
}
      • ujjwal78 + 1 comment

        Ah,thanks I got it.. It's all about algorithmic complexity.

        • gary_l_hewitt + 0 comments

          This is O(n) where n is the length of the queries array. It is independent of the length of the input array, since it grabs the input array elements by index and only has to loop through the queries to grab each one.