Skip to content

Latest commit

 

History

History
106 lines (87 loc) · 2.41 KB

File metadata and controls

106 lines (87 loc) · 2.41 KB

Merge Sort

  • It is a divide and conquer algorithm.
  • Divide the array into two halves, recursively sort the two halves and then merge the two sorted halves.
  • If array becomes empty or has only one element, then it is already sorted.
  • Recurrence relation for time complexity: T(n) = 2T(n/2) + θ(n)
  • TC: O(n log n)
  • SC: O(n)

Code

void merge(vector<int>& a, int l, int mid, int r)
{
    int n1 = mid - l + 1;
    int n2 = r - mid;
    vector<int> left(n1), right(n2);
    for (int i = 0; i < n1; i++) {
        left[i] = a[l + i];
    }
    for (int i = 0; i < n2; i++) {
        right[i] = a[mid + 1 + i];
    }

    int i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {
        if (left[i] <= right[j]) {
            a[k] = left[i];
            i++;
        } else {
            a[k] = right[j];
            j++;
        }
        k++;
    }
    while (i < n1) {
        a[k] = left[i];
        i++;
        k++;
    }
    while (j < n2) {
        a[k] = right[j];
        j++;
        k++;
    }
    // delete left, right;
    // left.clear();
    // right.clear();
}

void mergeSort(vector<int>& a, int l, int r)
{
    if (l >= r)
        return;

    int mid = l + (r - l) / 2;
    mergeSort(a, l, mid);
    mergeSort(a, mid + 1, r);
    merge(a, l, mid, r);
}

Analysis

  • Algorithmic Paradigm: Divide and Conquer
  • Is In-place Algorithm: No
  • Is Stable: Yes

Uses/Applications:

  • Sorting 2 linked list in O(n log n) time.
  • Inversion count problem.
  • External sorting.

Drawbacks

  • Slower for smaller task compared to other sorting algorithms.
  • Requires additional memory.
  • Goes through the whole process even of array is sorted.

Iterative Merge Sort

  • merge function remains same.
  • Read full article on GFG.
  • Time and space complexity is same as recursive solution but it does not uses recursive stack.

Code

void mergeSortIterative(vector<int>& a, int n)
{
    int currSize, left, right, mid;
    for (currSize = 1; currSize < n; currSize = 2 * currSize) {
        for (left = 0; left < n; left += 2 * currSize) {
            mid = min(left + currSize - 1, n - 1);
            right = min(left + 2 * currSize - 1, n - 1);

            merge(a, left, mid, right);
        }
    }
}

Reference