Merge Without Extra Space | Practice | GeeksforGeeks

Given two sorted arrays arr1[] and arr2[] of sizes n and m in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements. 
 
Example 1:
Input: 
n = 4, arr1[] = [1 3 5 7] 
m = 5, arr2[] = [0 2 6 8 9]
Output: 
arr1[] = [0 1 2 3]
arr2[] = [5 6 7 8 9]
Explanation:
After merging the two 
non-decreasing arrays, we get, 
0 1 2 3 5 6 7 8 9.
Example 2:
Input: 
n = 2, arr1[] = [10, 12] 
m = 3, arr2[] = [5 18 20]
Output: 
arr1[] = [5 10]
arr2[] = [12 18 20]
Explanation:
After merging two sorted arrays 
we get 5 10 12 18 20.



Your Task:
You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, n and m as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated.
 
Expected Time Complexity:  O((n+m) log(n+m))
Expected Auxilliary Space: O(1)
 
Constraints:
1 <= n, m <= 5*104
0 <= arr1i, arr2i <= 107
0
bandarusivasai3 days ago
class Solution
{
    //Function to merge the arrays.
    public static void merge(long arr1[], long arr2[], int n, int m) 
    {
        // code here 
        for(int i=0;i<n;i++)
        {
            if(arr1[i]>arr2[0])
            {
                long t=arr1[i];
                arr1[i]=arr2[0];
                arr2[0]=t;
                findmin(arr2);
            }
        }
        Arrays.sort(arr1);
        Arrays.sort(arr2);
    }
    public static void findmin(long arr[])
    {
        long min=arr[0];
        int index=0;
        for(int i=0;i<arr.length;i++)
        {
            if(arr[i]<min)
            {
                min=arr[i];
                index=i;
            }
        }
        long t=arr[index];
        arr[index]=arr[0];
        arr[0]=min;
    }
}
0
anilcangulkaya76 days ago
idk what is wrong in this code returns: 1 3 5 7 0 2 6 8 9 :
static int comp(const void* a, const void* b){   return *(int*)&a > *(int*)&b;}
void merge(int arr1[], int arr2[], int n, int m){    //code here   int* temp = malloc(n * 4 + m * 4);      memcpy(temp, arr1, n * 4);   memcpy(temp + n, arr2, m * 4);      qsort(temp, n + m, 4, comp);      memcpy(arr1, temp, n * 4);   memcpy(arr2, temp + n, m * 4);      free(temp);}
+1
navneet05082002kuswaha2 weeks ago
Striver Gap Method
import mathclass Solution:      #Function to merge the arrays.   def merge(self,arr1,arr2,n,m):                     gap = math.ceil((n+m)/2)              while gap:                      i = 0           j = gap                      while j < (n+m):                              if j < n and arr1[i]>arr1[j]:                   arr1[i],arr1[j] = arr1[j],arr1[i]                                  elif j>=n and i<n and arr1[i]>arr2[j-n]:                   arr1[i],arr2[j-n] = arr2[j-n],arr1[i]                                  elif j>=n and i>=n and arr2[i-n]>arr2[j-n]:                   arr2[i-n],arr2[j-n] = arr2[j-n],arr2[i-n]                              j += 1               i += 1                              if gap == 1:               gap = 0           else:               gap = math.ceil(gap/2) 
-2
haulya2 weeks ago
        void merge(long long arr1[], long long arr2[], int n, int m)        {            // code here            int i=n-1;           int j = 0;                      while(i >= 0 && j < m){               if(arr2[j] < arr1[i]){                   swap(arr2[j], arr1[i]);                   i--;               }               j++;           }                      sort(arr1, arr1+n);           sort(arr2, arr2+m);       }
0
cshubham4392 weeks ago
 int i = n-1;
        int j = 0;
        while(i>=0 && j<m){
            if(arr1[i]>arr2[j]){
                long temp = arr1[i];
                arr1[i] = arr2[j];
                arr2[j] = temp;
            }
            i--;
            j++;
        }
        Arrays.sort(arr1);
        Arrays.sort(arr2);
+1
shubham211019972 weeks ago
simplest solution
void merge(long long arr1[], long long arr2[], int n, int m) 
       { 
          int i=n-1,j=0;
   while(i>=0 && j<m){
       if(arr1[i]>arr2[j]){
           int temp=arr1[i];
           arr1[i]=arr2[j];
           arr2[j]=temp;
       }
       i--;
       j++;
   }
   sort(arr1,arr1+n);
   sort(arr2,arr2+m);
       } 
0
nagasivakrishnav2 weeks ago
Python code very short:
 
def merge(self,arr1,arr2,n,m):       arr1.extend(arr2)       temparr = arr1.copy()       temparr.sort()       arr1.clear()       arr1.extend(temparr[:n])       arr2.clear()       temparr.sort(reverse=True)       arr2.extend(sorted(temparr[:m]))
 
0
artistdarkangel3 weeks ago
BEST JAVA code. 
 
class Solution{  public static void merge(long arr1[], long arr2[], int n, int m)   {      int i=n-1;      int j=0;      while(i>=0 && j<m)      {          if(arr1[i]>arr2[j])          {              long temp=arr1[i];              arr1[i]=arr2[j];              arr2[j]=temp;          }          else           break;          i--;          j++;      }      Arrays.sort(arr1);      Arrays.sort(arr2);  }}
 
0
sagrikasoni3 weeks ago
JAVA SOLUTION
class Solution
{
    //Function to merge the arrays.
    public static void merge(long arr1[], long arr2[], int n, int m) 
    {
      for(int i =0;i<n;i++){
          if(arr1[i]>arr2[0]){
             long temp = arr1[i];
              arr1[i] = arr2[0];
              arr2[0] = temp;
          
       int k =0;
         long first = arr2[0];
         for(k =1;k<m && arr2[k]<first;k++){
             arr2[k-1] = arr2[k];
         }
         arr2[k-1] = first;
      }
    }
}
}
0
geek_deepx4 weeks ago
its showing TLE at 236th test case
my concept was to put smallest element of both the arrays at arr1[i] so i  compared arr1[i] with arr2[0] , because at every step i was sorting arr2 so there will be always smallest element of arr2 at arr2[0].
 
void merge(long long arr1[], long long arr2[], int n, int m)        {            for(int i = 0; i < n; i++) {               if(arr2[0] < arr1[i]){                   swap(arr2[0], arr1[i]);               }               sort(arr2, arr2+m);           }        }
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