Find three closest elements from given three sorted arrays

3.5

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
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
24 from A, 22 from B and 23 from C

We strongly recommend you to minimize your browser and try this yourself first.

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. Time complexity of this solution is O(n3)

A Better Solution is to us 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.

// C++ program to find 3 elements such that max(abs(A[i]-B[j]), abs(B[j]-
// C[k]), abs(C[k]-A[i])) is minimized.
#include<bits/stdc++.h>
using namespace std;

void findClosest(int A[], int B[], int C[], int p, int q, int r)
{

    int diff = INT_MAX;  // Initialize min diff

    // Initialize result
    int res_i =0, res_j = 0, res_k = 0;

    // Traverse arrays
    int i=0,j=0,k=0;
    while (i < p && j < q && k < r)
    {
        // Find minimum and maximum of current three elements
        int minimum = min(A[i], min(B[j], C[k]));
        int maximum = max(A[i], max(B[j], C[k]));

        // Update result if current diff is less than the min
        // diff so far
        if (maximum-minimum < diff)
        {
             res_i = i, res_j = j, res_k = k;
             diff = maximum - minimum;
        }

        // We can't get less than 0 as values are absolute
        if (diff == 0) break;

        // Increment index of array with smallest value
        if (A[i] == minimum) i++;
        else if (B[j] == minimum) j++;
        else k++;
    }

    // Print result
    cout << A[res_i] << " " << B[res_j] << " " << C[res_k];
}

// Driver program
int main()
{
    int A[] = {1, 4, 10};
    int B[] = {2, 15, 20};
    int C[] = {10, 12};

    int p = sizeof A / sizeof A[0];
    int q = sizeof B / sizeof B[0];
    int r = sizeof C / sizeof C[0];

    findClosest(A, B, C, p, q, r);
    return 0;
}

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.

Thanks to Gaurav Ahirwar for suggesting above solutions.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



3.5 Average Difficulty : 3.5/5.0
Based on 28 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.