Find three closest elements from given three sorted arrays

Given three sorted arrays A[], B[] and C[], find 3 elements i, j and k from A, B and C respectively such that max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) is minimized. Here abs() indicates absolute value.

Example : 

Input : A[] = {1, 4, 10}
            B[] = {2, 15, 20}
            C[] = {10, 12}

Output: 10 15 10
Explanation: 10 from A, 15 from B and 10 from C

Input: A[] = {20, 24, 100}
           B[] = {2, 19, 22, 79, 800}
          C[] = {10, 12, 23, 24, 119}
Output: 24 22 23
Explanation: 24 from A, 22 from B and 23 from C

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A Simple Solution is to run three nested loops to consider all triplets from A, B and C. Compute the value of max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) for every triplet and return minimum of all values.

Steps to implement-

• Declared three variables a,b, and c to store final answers
• Initialized a variable “ans” with the Maximum value
• We will store a minimum of max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) in “ans” variable
• Run three nested loops where each loop is for each array
• From that loop if max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) is less than a minimum of max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) present in “ans”
  • Then, update “ans” with new max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) and that three variable a,b,c with three elements of those arrays
• In the last print value present in a,b,c

Code-

Output-

10 15 10

Time complexity :O(N³),because of three nested loops
Auxiliary space: O(1),because no extra space has been used

A Better Solution is to use Binary Search. 
1) Iterate over all elements of A[], 
      a) Binary search for element just smaller than or equal to in B[] and C[], and note the difference. 
2) Repeat step 1 for B[] and C[]. 
3) Return overall minimum.
Time complexity of this solution is O(nLogn)

Efficient Solution Let ‘p’ be size of A[], ‘q’ be size of B[] and ‘r’ be size of C[]  

1)   Start with i=0, j=0 and k=0 (Three index variables for A,
                                  B and C respectively)
//  p, q and r are sizes of A[], B[] and C[] respectively.
2)   Do following while i < p and j < q and k < r
    a) Find min and maximum of A[i], B[j] and C[k]
    b) Compute diff = max(X, Y, Z) - min(A[i], B[j], C[k]).
    c) If new result is less than current result, change 
       it to the new result.
    d) Increment the pointer of the array which contains 
       the minimum.

Note that we increment the pointer of the array which has the minimum because our goal is to decrease the difference. Increasing the maximum pointer increases the difference. Increase the second maximum pointer can potentially increase the difference. 

Output: 

10 15 10

Time complexity of this solution is O(p + q + r) where p, q and r are sizes of A[], B[] and C[] respectively.
Auxiliary space: O(1) as constant space is required.

Approach 2: Using Binary Search:

Another approach to solve this problem can be to use binary search along with two pointers.

First, sort all the three arrays A, B, and C. Then, we take three pointers, one for each array. For each i, j, k combination, we calculate the maximum difference using the absolute value formula given in the problem. If the current maximum difference is less than the minimum difference found so far, then we update our result.

Next, we move our pointers based on the value of the maximum element among the current i, j, k pointers. We increment the pointer of the array with the smallest maximum element, hoping to find a smaller difference.

The time complexity of this approach will be O(nlogn) due to sorting, where n is the size of the largest array.

Here’s the code for this approach:

OUTPUT:

10 15 10

Time complexity :O(NlogN) 
Auxiliary space: O(1) as constant space is required.

//Thanks to Gaurav Ahirwar for suggesting the above solutions.
 

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