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  1. Practice
  2. Data Structures
  3. Arrays
  4. Left Rotation
  5. Discussions

Left Rotation

  • corey_t_ferguson + 2 comments

    To solve this, I've shifted the elements in the array in O(n) time and O(n) memory (vs O(2n)). Not as easy on the eyes, though :(

    function processData(input) {
  // get `n` and `d` from input
  const lines = input.split('\n');
  const firstLine = lines[0].split(' ').map(Number);
  const n = firstLine[0];
  const d = firstLine[1];

  // process each line
  lines.slice(1, lines.length).forEach(line => {
    // no need to shift in these cases
    if (n === 1 || d === n) {
      console.info(line);
    } else {
      // shift digits
      const a = line.split(' ').map(Number);
      let lastLastItem = null;
      let count = 0;
      let i = 0;
      while (count < n) {
        i++;
        const start = i;
        let j = start;
        do {
          count++;
          let lastItem = lastLastItem;
          lastLastItem = a[j];
          a[j] = lastItem;
          j = shiftLeft(n, d, j);
        } while (j !== start);
        a[start] = lastLastItem;
      }
      console.info(a.reduce((acc, value) => {
        return acc+' '+value;
      }));
    }
  });
}

/**
 * @param {Number} n total number of elements
 * @param {Number} d number of shifts to left
 * @param {Number} i index to begin shifting from
 * @returns {Number} new index after shifting to left
 */
function shiftLeft(n, d, i) {
  return (n-d+i)%n;
}

    I'm ignoring the processing time on input string manipulation and such. This examples assumes they gave us an existing array to manipulate.

    • vjindal23 + 0 comments

      Hey! Could you include comments in your code please? I use C and dont kow other language as of now but would like to learn from your solution O(n) is preety nice

    • siregar_jaswir + 1 comment

      i think your solutions is O(n^2) for the worst case.

      • corey_t_ferguson + 0 comments

        It looks like it, because of the nested loops. But it'll break out of all loops after n loops.