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

Circular Array Rotation

  • polmki99 + 7 comments

    What's the space and time efficiency for this code?

    n, k, m = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))
k %= n
arr = arr[-k:] + arr[:-k]
for i in range(m):
    print(arr[int(input().strip())])
    • hgnisrednivrug + 0 comments

      wow!!! fast and easy way to done the problem..

    • Pr0gr3550 + 0 comments
      [deleted]
    • franchb1 + 0 comments

      Thank you!

      Now I know %=

    • amcavinue + 0 comments

      Really great solution!

    • nstoddar + 1 comment

      Thanks polmki99! Very clever implementation.

      Here it is in Javascript for those interested.

      function processData(input) {
    //Enter your code here
    var inputList = input.split('\n');
    var details = inputList[0].split(" ");
    var arrayIdx = inputList.slice(2);
    var targetArray = inputList[1].split(" ");
    
    var k = details[1]%details[0];
    
    targetArray = [].concat(targetArray.slice(-k),targetArray.slice(0,-k));
    
    arrayIdx.map(function(elem){return console.log(targetArray[elem]);});
}
      • 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.

    • ryanlough + 0 comments
      [deleted]
    • mail2karthk07 + 0 comments

      how about my solution is it good..

      n,k,q = list(map(int, input().strip().split()))
a = list(map(int, input().strip().split()))
a = a[n-(k%n):]+a[0:n-(k%n)]
[print(a[int(input())]) for _ in range(q)]