{"componentChunkName":"component---src-templates-blog-post-jsx","path":"/blogs/mergeSort","result":{"data":{"blog":{"frontmatter":{"title":"MERGE SORT","thumbnail":"blog61","date":"May 2, 2021","dsaCppCodeFile":"https://drive.google.com/file/d/1NeaBoa0c-OvYpi-qgL-3JeRvmJdIUmnT/view?usp=sharing"},"excerpt":"
\n

Introduction

\n
\n

\n Merge Sort divides the array(or given data structure) into two\n halves, and again calls itself recursively until single\n element is left and then merges the two sorted halves. It\n requires a temporary array to merge and store the sorted\n array.\n

\n

\n This algorithm falls under Divide & Conquer Technique which is\n divides the problem, finds the solution to smaller problem and\n then merges these to get the solution to the original problem.\n Let's look at the working of this algorithm.\n

\n
\n
\n
\n

Working Procedure

\n

\n Do refer the code available the end of this section to\n understand the following theory.\n

\n
    \n
  1. \n Consider an array,arr of n elements. Let left be the starting\n index and right be the ending index.\n
    2. \n
  2. \n Merge Sort: \n
      \n
    1. \n Recursion calls will be made if there exist more than one\n element in the array.\n
      2. \n
    2. \n Find the mid point of the array, mid=(left+right)/2.\n
      3. \n
    3. Make the first recursive call from left to mid.
      4. \n
    4. Make the second recursive call from mid+1 to right.
      5. \n
    5. \n Now merge this array from index left to right using\n merge function.\n
      6. \n
    \n
    3. \n
  3. \n Merge Function:\n
      \n
    1. \n Let i=left be the start index for left subarray, j=mid+1\n be the start index for right subarray and k=left be the\n start index for temporary array.\n
      2. \n
    2. \n Loop until i<=left and r<=right(That is either of the\n subarray is looped completely).\n
      3. \n
        \n
      1. \n If arr[i] is lesser than arr[j](if the current element\n in left subarray is lesser than current element in right\n subarray) then copy arr[i] to temp[k] and increment i,\n k.\n
        2. \n
      2. Else copy arr[j] to temp[k] and increment j, k.
        3. \n
      \n
    3. \n Copy the remaining elements of left subarray to temporary\n array(if there are any).\n
      4. \n
    4. \n Copy the remaining elements of right subarray to temporary\n array(if there are any).\n
      5. \n
    5. \n Finally copy the elements from temporary array to our\n required array(temp->arr).\n
      6. \n
    \n
    4. \n
\n
\n
\n

Time Complexity

\n
\n

\n PS: We give random inputs because the no. of input will\n generally be 100 or 1000 to find the time complexity and\n manually giving so many inputs is cumbersome work.\n

\n

\n There exists two recursive calls in which the elements gets\n halved in each call. And the transfer of elements from\n arr->temp and at last from temp->arr will effect the time\n complexity. So declare a global variable count, increment this\n variable in all the loops present in the merge function\n (within the loops at step 3.2, 3.3, 3.4 & 3.5 in the above\n working procedure).\n

\n \n \n \n \"Merge\n \n \n

\n Time Complexity: O(nlog(n))\n

\n
\n
\n","html":"
\n

Introduction

\n
\n

\n Merge Sort divides the array(or given data structure) into two\n halves, and again calls itself recursively until single\n element is left and then merges the two sorted halves. It\n requires a temporary array to merge and store the sorted\n array.\n

\n

\n This algorithm falls under Divide & Conquer Technique which is\n divides the problem, finds the solution to smaller problem and\n then merges these to get the solution to the original problem.\n Let's look at the working of this algorithm.\n

\n
\n
\n
\n

Working Procedure

\n

\n Do refer the code available the end of this section to\n understand the following theory.\n

\n
    \n
  1. \n Consider an array,arr of n elements. Let left be the starting\n index and right be the ending index.\n
    2. \n
  2. \n Merge Sort: \n
      \n
    1. \n Recursion calls will be made if there exist more than one\n element in the array.\n
      2. \n
    2. \n Find the mid point of the array, mid=(left+right)/2.\n
      3. \n
    3. Make the first recursive call from left to mid.
      4. \n
    4. Make the second recursive call from mid+1 to right.
      5. \n
    5. \n Now merge this array from index left to right using\n merge function.\n
      6. \n
    \n
    3. \n
  3. \n Merge Function:\n
      \n
    1. \n Let i=left be the start index for left subarray, j=mid+1\n be the start index for right subarray and k=left be the\n start index for temporary array.\n
      2. \n
    2. \n Loop until i<=left and r<=right(That is either of the\n subarray is looped completely).\n
      3. \n
        \n
      1. \n If arr[i] is lesser than arr[j](if the current element\n in left subarray is lesser than current element in right\n subarray) then copy arr[i] to temp[k] and increment i,\n k.\n
        2. \n
      2. Else copy arr[j] to temp[k] and increment j, k.
        3. \n
      \n
    3. \n Copy the remaining elements of left subarray to temporary\n array(if there are any).\n
      4. \n
    4. \n Copy the remaining elements of right subarray to temporary\n array(if there are any).\n
      5. \n
    5. \n Finally copy the elements from temporary array to our\n required array(temp->arr).\n
      6. \n
    \n
    4. \n
\n
\n
\n

Time Complexity

\n
\n

\n PS: We give random inputs because the no. of input will\n generally be 100 or 1000 to find the time complexity and\n manually giving so many inputs is cumbersome work.\n

\n

\n There exists two recursive calls in which the elements gets\n halved in each call. And the transfer of elements from\n arr->temp and at last from temp->arr will effect the time\n complexity. So declare a global variable count, increment this\n variable in all the loops present in the merge function\n (within the loops at step 3.2, 3.3, 3.4 & 3.5 in the above\n working procedure).\n

\n \n \n \n \"Merge\n \n \n

\n Time Complexity: O(nlog(n))\n

\n
\n
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