包含矩阵每一行元素的数组的相邻元素之间的最小差异

给定一个N行M列的矩阵，任务是找到大小为N的数组的两个相邻元素中的任何一个之间的最小绝对差，该数组是通过从矩阵的每一行中选取一个元素来创建的。请注意，从第 1 行选取的元素将变为 arr[0]，从第 2 行选取的元素将变为 arr[1]，依此类推。

例子：

Input :  N = 2, M = 2
m[2][2] = { 8, 2,
            6, 8 }
Output : 0.
Picking 8 from row 1 and picking 8 from row 2, we create an array { 8, 8 } and minimum
difference between any of adjacent element is 0.

Input :  N = 3, M = 3
m[3][3] = { 1, 2, 3
            4, 5, 6 
            7, 8, 9 }
Output : 1.

这个想法是单独对所有行进行排序，然后进行二分搜索以找到每个元素的下一行中最接近的元素。
要以有效的方式执行此操作，请对矩阵的每一行进行排序。从矩阵的第 1 行到第 N – 1 行，对于矩阵中当前行的每个元素 m[i][j]，找出下一行中大于或等于当前元素的最小元素，例如 p和小于当前元素的最大元素，比如 q。这可以使用二分搜索来完成。最后，从 p 和 q 中找出当前元素的差值的最小值并更新变量。

下面是这种方法的实现：

C++

// C++ program to find the minimum absolute difference
// between any of the adjacent elements of an array
// which is created by picking one element from each
// row of the matrix.
#include
using namespace std;
#define R 2
#define C 2
 
// Return smallest element greater than or equal
// to the current element.
int bsearch(int low, int high, int n, int arr[])
{
    int mid = (low + high)/2;
 
    if(low <= high)
    {
        if(arr[mid] < n)
            return bsearch(mid +1, high, n, arr);
        return bsearch(low, mid - 1, n, arr);
    }
 
    return low;
}
 
// Return the minimum absolute difference adjacent
// elements of array
int mindiff(int arr[R][C], int n, int m)
{
    // Sort each row of the matrix.
    for (int i = 0; i < n; i++)
        sort(arr[i], arr[i] + m);
 
    int ans = INT_MAX;
 
    // For each matrix element
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < m; j++)
        {
            // Search smallest element in the next row which
            // is greater than or equal to the current element
            int p = bsearch(0, m-1, arr[i][j], arr[i + 1]);
            ans = min(ans, abs(arr[i + 1][p] - arr[i][j]));
 
            // largest element which is smaller than the current
            // element in the next row must be just before
            // smallest element which is greater than or equal
            // to the current element because rows are sorted.
            if (p-1 >= 0)
                ans = min(ans, abs(arr[i + 1][p - 1] - arr[i][j]));
        }
    }
    return ans;
}
 
// Driven Program
int main()
{
    int m[R][C] =
    {
        8, 5,
        6, 8,
    };
 
    cout << mindiff(m, R, C) << endl;
    return 0;
}

Java

// Java program to find the minimum
// absolute difference between any
// of the adjacent elements of an
// array which is created by picking
// one element from each row of the matrix
import java.util.Arrays;
class GFG
{
static final int R=2;
static final int C=2;
 
// Return smallest element greater than
// or equal to the current element.
static int bsearch(int low, int high, int n, int arr[])
{
    int mid = (low + high)/2;
 
    if(low <= high)
    {
        if(arr[mid] < n)
            return bsearch(mid +1, high, n, arr);
        return bsearch(low, mid - 1, n, arr);
    }
 
    return low;
}
 
// Return the minimum absolute difference adjacent
// elements of array
static int mindiff(int arr[][], int n, int m)
{
 
    // Sort each row of the matrix.
    for (int i = 0; i < n; i++)
        Arrays.sort(arr[i]);
 
    int ans = [phone];
 
    // For each matrix element
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < m; j++)
        {
 
        // Search smallest element in the
        // next row which is greater than
        // or equal to the current element
        int p = bsearch(0, m-1, arr[i][j], arr[i + 1]);
        ans = Math.min(ans, Math.abs(arr[i + 1][p] - arr[i][j]));
 
        // largest element which is smaller than the current
        // element in the next row must be just before
        // smallest element which is greater than or equal
        // to the current element because rows are sorted.
        if (p-1 >= 0)
            ans = Math.min(ans,
                        Math.abs(arr[i + 1][p - 1] - arr[i][j]));
        }
    }
    return ans;
}
 
    //Driver code
public static void main (String[] args)
{
    int m[][] ={{8, 5},
                {6, 8}};
 
    System.out.println(mindiff(m, R, C));
}
}
//This code is contributed by Anant Agarwal.

Python

# Python program to find the minimum absolute difference
# between any of the adjacent elements of an array
# which is created by picking one element from each
# row of the matrix.
# R 2
# C 2
  
# Return smallest element greater than or equal
# to the current element.
def bsearch(low, high, n, arr):
    mid = (low + high)/2
  
    if(low <= high):
        if(arr[mid] < n):
            return bsearch(mid +1, high, n, arr);
        return bsearch(low, mid - 1, n, arr);
  
    return low;
  
# Return the minimum absolute difference adjacent
# elements of array
def mindiff(arr, n, m):
 
    # arr = [0 for i in range(R)][for j in range(C)]
    # Sort each row of the matrix.
    for i in range(n):
        sorted(arr)
  
    ans = 2147483647
  
    # For each matrix element
    for i in range(n-1):
        for j in range(m):
            # Search smallest element in the next row which
            # is greater than or equal to the current element
            p = bsearch(0, m-1, arr[i][j], arr[i + 1])
            ans = min(ans, abs(arr[i + 1][p] - arr[i][j]))
  
            # largest element which is smaller than the current
            # element in the next row must be just before
            # smallest element which is greater than or equal
            # to the current element because rows are sorted.
            if (p-1 >= 0):
                ans = min(ans, abs(arr[i + 1][p - 1] - arr[i][j]))
    return ans;
  
# Driver Program
m =[8, 5], [6, 8]
print mindiff(m, 2, 2)
 
# This code is contributed by Afzal

C#

// C# program to find the minimum
// absolute difference between any
// of the adjacent elements of an
// array which is created by picking
// one element from each row of the matrix
using System;
 
class GFG
{
     
static public int R=2;
static public int C=2;
 
// Return smallest element greater than
// or equal to the current element.
static int bsearch(int low, int high, int n, int []arr)
{
    int mid = (low + high)/2;
 
    if(low <= high)
    {
        if(arr[mid] < n)
            return bsearch(mid +1, high, n, arr);
        return bsearch(low, mid - 1, n, arr);
    }
 
    return low;
}
public static int[] GetRow(int[,] matrix, int row)
{
    var rowLength = matrix.GetLength(1);
    var rowVector = new int[rowLength];
 
    for (var i = 0; i < rowLength; i++)
    rowVector[i] = matrix[row, i];
 
    return rowVector;
}
// Return the minimum absolute difference adjacent
// elements of array
static int mindiff(int [,]arr, int n, int m)
{
 
    // Sort each row of the matrix.
    for (int i = 0; i < n; i++)
        Array.Sort(GetRow(arr,i));
 
    int ans = [phone];
 
    // For each matrix element
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < m; j++)
        {
 
        // Search smallest element in the
        // next row which is greater than
        // or equal to the current element
        int p = bsearch(0, m-1, arr[i,j], GetRow(arr,i+1));
        ans = Math.Min(ans, Math.Abs(arr[i + 1,p] - arr[i,j]));
 
        // largest element which is smaller than the current
        // element in the next row must be just before
        // smallest element which is greater than or equal
        // to the current element because rows are sorted.
        if (p-1 >= 0)
            ans = Math.Min(ans,
                        Math.Abs(arr[i + 1,p - 1] - arr[i,j]));
        }
    }
    return ans;
}
 
// Driver code
public static void Main (String[] args)
{
    int [,]m ={{8, 5},
                {6, 8}};
 
    Console.WriteLine(mindiff(m, R, C));
}
}
 
// This code is contributed by 29AjayKumar

PHP

= 0)
                $ans = min($ans, abs($arr[$i + 1][$p - 1] -
                                     $arr[$i][$j]));
        }
    }
    return $ans;
}
 
// Driver Code
$m = array(8, 5, 6, 8);
 
echo mindiff($m, $R, $C), "\n";
 
// This code is contributed by Sach_Code
?>

Javascript

 
// JavaScript program to find the minimum
// absolute difference between any
// of the adjacent elements of an
// array which is created by picking
// one element from each row of the matrix.
let R = 2;
let C = 2;
   
// Return smallest element greater than
// or equal to the current element.
function bsearch(low, high, n, arr)
{
    let mid = (low + high) / 2;
   
    if (low <= high)
    {
        if (arr[mid] < n)
            return bsearch(mid + 1,
                           high, n, arr);
             
        return bsearch(low, mid - 1,
                       n, arr);
    }
    return low;
}
   
// Return the minimum absolute difference
// adjacent elements of array
function mindiff(arr, n, m)
{
     
    // Sort each row of the matrix.
    for(let i = 0; i < n; i++)
        arr.sort();
   
    let ans = [phone];
   
    // For each matrix element
    for(let i = 0; i < n - 1; i++)
    {
        for(let j = 0; j < m; j++)
        {
             
            // Search smallest element in the
            // next row which is greater than
            // or equal to the current element
            let p = bsearch(0, m - 1, arr[i][j],
                                      arr[i + 1]);
            ans = Math.min(ans,
                  Math.abs(arr[i + 1][p] -
                           arr[i][j]));
       
            // largest element which is smaller
            // than the current element in the
            // next row must be just before smallest
            // element which is greater than or equal
            // to the current element because
            // rows are sorted.
            if (p - 1 >= 0)
                ans = Math.min(ans,
                      Math.abs(arr[i + 1][p - 1] -
                               arr[i][j]));
        }
    }
    return ans;
}
  
// Driver Code
let m = [ [ 8, 5 ],
          [ 6, 8 ] ];
 
document.write(mindiff(m, R, C));
 
// This code is contributed by code_hunt
 

输出：

0

时间复杂度： O(N*M*logM)。

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