给定长度为N且范围为D的范围arr []的数组，任务是找到最小数量，以改变数量D来使D处于给定数组的每个范围内。此处，范围由两个整数[start，end]组成，并且如果start≤D≤end ，则D位于范围内。
例子：
Input: N = 3, D = 3
arr[] = {{0, 7}, {2, 14}, {4, 6}}
Output: 1
Explanation:
Here if we increment D by 1 then it will be inside of every range that is start ≤ D ≤ end.
Input: N = 2, D = 2
arr[] = {{1, 2}, {3, 2}}
Output: 0
Explanation:
Here D = 2 which is already inside of every range that is start ≤ D ≤ end.
应用方法：
- 遍历范围数组，找到start [i]的最大值和end [i]的最小值，其中最终的max_start和min_end将显示给定范围数组的公共范围。
- 检查D是否已在max_start和min_end中。
- 如果D已经在范围内，则最小差为0。
- 否则，从max_start和min_end中找到最接近D的值，即
min(abs(D - max_start), abs(D - min_end))
下面是上述方法的代码：
C++
// C++ implementation find the minimum
// difference in the number D such that
// D is inside of every range
#include
using namespace std;
// Function to find the minimum
// difference in the number D such that
// D is inside of every range
void findMinimumOperation(int n, int d,
int arrays[3][2]){
int cnt = 0;
int first = INT_MIN, end = INT_MAX;
// Loop to find the common range out
// of the given array of ranges.
while (n--) {
// Storing the start and end index
int arr[2] = { arrays[cnt][0],
arrays[cnt][1] };
// Sorting the range
sort(arr, arr + 2);
// Finding the maximum starting
// value of common segment
first = max(first, arr[0]);
// Finding the minimum ending
// value of common segment
end = min(end, arr[1]);
cnt++;
}
// If there is no common segment
if (first > end)
cout << "-1";
else {
// If the given number is between
// the computed common range.
if (d >= first && d <= end) {
cout << "0";
}
// Finding the minimum distance
else
cout << min(abs(first - d),
abs(d - end));
}
}
// Driver Code
int main()
{
int n = 3, d = 3;
// Given array of ranges
int arrays[3][2] = {
{ 0, 7 },
{ 2, 14 },
{ 4, 6 }
};
findMinimumOperation(n, d, arrays);
} Java
// Java implementation find the minimum
// difference in the number D such that
// D is inside of every range
import java.util.*;
class GFG
{
// Function to find the minimum
// difference in the number D such that
// D is inside of every range
static void findMinimumOperation(int n, int d,
int arrays[][]){
int cnt = 0;
int first = Integer.MIN_VALUE, end = Integer.MAX_VALUE;
// Loop to find the common range out
// of the given array of ranges.
while (n > 0) {
// Storing the start and end index
int arr[] = { arrays[cnt][0],
arrays[cnt][1] };
// Sorting the range
Arrays.sort(arr);
// Finding the maximum starting
// value of common segment
first = Math.max(first, arr[0]);
// Finding the minimum ending
// value of common segment
end = Math.min(end, arr[1]);
cnt++;
n--;
}
// If there is no common segment
if (first > end)
System.out.print("-1");
else {
// If the given number is between
// the computed common range.
if (d >= first && d <= end) {
System.out.print("0");
}
// Finding the minimum distance
else
System.out.print(Math.min(Math.abs(first - d),
Math.abs(d - end)));
}
}
// Driver Code
public static void main (String []args)
{
int n = 3, d = 3;
// Given array of ranges
int arrays[][] = {
{ 0, 7 },
{ 2, 14 },
{ 4, 6 }
};
findMinimumOperation(n, d, arrays);
}
}
// This code is contributed by chitranayalPython3
# Python3 implementation find the minimum
# difference in the number D such that
# D is inside of every range
# Function to find the minimum
# difference in the number D such that
# D is inside of every range
def findMinimumOperation(n, d,arrays):
cnt = 0
first = -10**9
end = 10**9
# Loop to find the common range out
# of the given array of ranges.
while (n):
# Storing the start and end index
arr = [arrays[cnt][0],arrays[cnt][1]]
# Sorting the range
arr = sorted(arr)
# Finding the maximum starting
# value of common segment
first = max(first, arr[0])
# Finding the minimum ending
# value of common segment
end = min(end, arr[1])
cnt += 1
n -= 1
# If there is no common segment
if (first > end):
print("-1",end="")
else:
# If the given number is between
# the computed common range.
if (d >= first and d <= end):
print("0",end="")
# Finding the minimum distance
else:
print(min(abs(first - d),abs(d - end)),end="")
# Driver Code
if __name__ == '__main__':
n = 3
d = 3
# Given array of ranges
arrays=[[0, 7],
[2, 14],
[4, 6] ]
findMinimumOperation(n, d, arrays)
# This code is contributed by mohit kumar 29C#
// C# implementation find the minimum
// difference in the number D such that
// D is inside of every range
using System;
class GFG
{
// Function to find the minimum
// difference in the number D such that
// D is inside of every range
static void findMinimumOperation(int n, int d,
int [,]arrays){
int cnt = 0;
int first = int.MinValue, end = int.MaxValue;
// Loop to find the common range out
// of the given array of ranges.
while (n > 0) {
// Storing the start and end index
int []arr = { arrays[cnt, 0],
arrays[cnt, 1] };
// Sorting the range
Array.Sort(arr);
// Finding the maximum starting
// value of common segment
first = Math.Max(first, arr[0]);
// Finding the minimum ending
// value of common segment
end = Math.Min(end, arr[1]);
cnt++;
n--;
}
// If there is no common segment
if (first > end)
Console.Write("-1");
else {
// If the given number is between
// the computed common range.
if (d >= first && d <= end) {
Console.Write("0");
}
// Finding the minimum distance
else
Console.Write(Math.Min(Math.Abs(first - d),
Math.Abs(d - end)));
}
}
// Driver Code
public static void Main(String []args)
{
int n = 3, d = 3;
// Given array of ranges
int [,]arrays = {
{ 0, 7 },
{ 2, 14 },
{ 4, 6 }
};
findMinimumOperation(n, d, arrays);
}
}
// This code is contributed by PrinciRaj1992输出：
1