给定整数对的数组间隔,这些整数对表示大小为N的间隔的起点和终点。任务是从给定的间隔集中找到最小的非负整数,该整数是不重叠的数字。
输入约束:
例子:
Input: interval = {{0, 4}, {6, 8}, {2, 3}, {9, 18}}
Output: 5
Explanation:
The smallest non-negative integer which is non-overlapping to all set of the intervals is 5.
Input: interval = {{0, 14}, {86, 108}, {22, 30}, {5, 17}}
Output: 18
天真的方法:
- 创建一个大小为MAX的访问数组,并从开始到结束为每个间隔将所有值标记为true。
- 最后,从1迭代到MAX ,找到未被访问的最小值。
但是,如果间隔坐标不超过10 9 ,则此方法将不起作用。
时间复杂度: O(N 2 )
辅助空间: O(MAX)
高效方法:
- 无需从头到尾进行迭代,只需创建一个访问数组,并为每个范围标记vis [start] = 1和vis [end + 1] = -1即可。
- 取数组的前缀和。
- 然后遍历数组以找到值为0的第一个整数。
这是上述方法的实现:
C++
// C++ program to find the
// least non-overlapping number
// from a given set intervals
#include
using namespace std;
const int MAX = 1e5 + 5;
// function to find the smallest
// non-overlapping number
void find_missing(
vector > interval)
{
// create a visited array
vector vis(MAX);
for (int i = 0; i < interval.size(); ++i) {
int start = interval[i].first;
int end = interval[i].second;
vis[start]++;
vis[end + 1]--;
}
// find the first missing value
for (int i = 1; i < MAX; i++) {
vis[i] += vis[i - 1];
if (!vis[i]) {
cout << i << endl;
return;
}
}
}
// Driver function
int main()
{
vector > interval
= { { 0, 14 }, { 86, 108 },
{ 22, 30 }, { 5, 17 } };
find_missing(interval);
return 0;
}
Java
// Java program to find the
// least non-overlapping number
// from a given set intervals
class GFG{
static int MAX = (int) (1e5 + 5);
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// function to find the smallest
// non-overlapping number
static void find_missing(
pair[] interval)
{
// create a visited array
int [] vis = new int[MAX];
for (int i = 0; i < interval.length; ++i)
{
int start = interval[i].first;
int end = interval[i].second;
vis[start]++;
vis[end + 1]--;
}
// find the first missing value
for (int i = 1; i < MAX; i++) {
vis[i] += vis[i - 1];
if (vis[i]==0) {
System.out.print(i +"\n");
return;
}
}
}
// Driver function
public static void main(String[] args)
{
pair []interval = {new pair( 0, 14 ),
new pair( 86, 108 ),
new pair( 22, 30 ),
new pair( 5, 17 )};
find_missing(interval);
}
}
// This code is contributed by Rohit_ranjan
Python3
# Python3 program to find the
# least non-overlapping number
# from a given set intervals
MAX = int(1e5 + 5)
# Function to find the smallest
# non-overlapping number
def find_missing(interval):
# Create a visited array
vis = [0] * (MAX)
for i in range(len(interval)):
start = interval[i][0]
end = interval[i][1]
vis[start] += 1
vis[end + 1] -= 1
# Find the first missing value
for i in range(1, MAX):
vis[i] += vis[i - 1]
if (vis[i] == 0):
print(i)
return
# Driver code
interval = [ [ 0, 14 ], [ 86, 108 ],
[ 22, 30 ], [ 5, 17 ] ]
find_missing(interval)
# This code is contributed by divyeshrabadiya07
C#
// C# program to find the
// least non-overlapping number
// from a given set intervals
using System;
class GFG{
static int MAX = (int)(1e5 + 5);
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find the smallest
// non-overlapping number
static void find_missing(pair[] interval)
{
// Create a visited array
int [] vis = new int[MAX];
for(int i = 0; i < interval.Length; ++i)
{
int start = interval[i].first;
int end = interval[i].second;
vis[start]++;
vis[end + 1]--;
}
// Find the first missing value
for(int i = 1; i < MAX; i++)
{
vis[i] += vis[i - 1];
if (vis[i] == 0)
{
Console.Write(i + "\n");
return;
}
}
}
// Driver code
public static void Main(String[] args)
{
pair []interval = { new pair(0, 14),
new pair(86, 108),
new pair(22, 30),
new pair(5, 17) };
find_missing(interval);
}
}
// This code is contributed by Amit Katiyar
C++
// C++ program to find the
// least non-overlapping number
// from a given set intervals
#include
using namespace std;
// function to find the smallest
// non-overlapping number
void find_missing(
vector > interval)
{
// Sort the intervals based on their
// starting value
sort(interval.begin(), interval.end());
int mx = 0;
for (int i = 0; i < (int)interval.size(); ++i) {
// check if any missing vaue exist
if (interval[i].first > mx) {
cout << mx;
return;
}
else
mx = max(mx, interval[i].second + 1);
}
// finally print the missing value
cout << mx;
}
// Driver function
int main()
{
vector > interval
= { { 0, 14 }, { 86, 108 },
{ 22, 30 }, { 5, 17 } };
find_missing(interval);
return 0;
}
Java
// Java program to find the
// least non-overlapping number
// from a given set intervals
import java.util.*;
import java.io.*;
class GFG{
static class Pair implements Comparable
{
int start,end;
Pair(int s, int e)
{
start = s;
end = e;
}
public int compareTo(Pair p)
{
return this.start - p.start;
}
}
// Function to find the smallest
// non-overlapping number
static void findMissing(ArrayList interval)
{
// Sort the intervals based on their
// starting value
Collections.sort(interval);
int mx = 0;
for(int i = 0; i < interval.size(); ++i)
{
// Check if any missing vaue exist
if (interval.get(i).start > mx)
{
System.out.println(mx);
return;
}
else
mx = Math.max(mx, interval.get(i).end + 1);
}
// Finally print the missing value
System.out.println(mx);
}
// Driver code
public static void main(String []args)
{
ArrayList interval = new ArrayList<>();
interval.add(new Pair(0, 14));
interval.add(new Pair(86, 108));
interval.add(new Pair(22, 30));
interval.add(new Pair(5, 17));
findMissing(interval);
}
}
// This code is contributed by Ganeshchowdharysadanala
Python3
# Python3 program to find the
# least non-overlapping number
# from a given set intervals
# function to find the smallest
# non-overlapping number
def find_missing(interval):
# Sort the intervals based
# on their starting value
interval.sort()
mx = 0
for i in range (len(interval)):
# Check if any missing
# vaue exist
if (interval[i][0] > mx):
print (mx)
return
else:
mx = max(mx,
interval[i][1] + 1)
# Finally print the missing value
print (mx)
# Driver code
if __name__ == "__main__":
interval = [[0, 14], [86, 108],
[22, 30], [5, 17]]
find_missing(interval);
# This code is contributed by Chitranayal
C#
// C# program to find the
// least non-overlapping number
// from a given set intervals
using System;
using System.Collections.Generic;
class GFG{
class Pair : IComparable
{
public int start,end;
public Pair(int s, int e)
{
start = s;
end = e;
}
public int CompareTo(Pair p)
{
return this.start - p.start;
}
}
// Function to find the smallest
// non-overlapping number
static void findMissing(List interval)
{
// Sort the intervals based on their
// starting value
interval.Sort();
int mx = 0;
for(int i = 0; i < interval.Count; ++i)
{
// Check if any missing vaue exist
if (interval[i].start > mx)
{
Console.WriteLine(mx);
return;
}
else
mx = Math.Max(mx, interval[i].end + 1);
}
// Finally print the missing value
Console.WriteLine(mx);
}
// Driver code
public static void Main(String []args)
{
List interval = new List();
interval.Add(new Pair(0, 14));
interval.Add(new Pair(86, 108));
interval.Add(new Pair(22, 30));
interval.Add(new Pair(5, 17));
findMissing(interval);
}
}
// This code is contributed by shikhasingrajput
18
时间复杂度: O(N)
辅助空间: O(MAX)
但是,如果间隔坐标最大为10 9 ,则此方法也将不起作用。
高效方法:
- 按其起始坐标和每个下一个范围对范围进行排序。
- 检查起点是否大于到目前为止遇到的最大终点坐标,然后可以找到一个丢失的数字,该数字将为previous_max + 1 。
Illustration:
Consider the following example:
interval[][] = { { 0, 14 }, { 86, 108 }, { 22, 30 }, { 5, 17 } };
After sorting, interval[][] = { { 0, 14 }, { 5, 17 }, { 22, 30 }, { 86, 108 }};
Initial mx = 0 and after considering first interval mx = max(0, 15) = 15
Since mx = 15 and 15 > 5 so after considering second interval mx = max(15, 18) = 18
now 18 < 22 so 18 is least non-overlapping number.
这是上述方法的实现:
C++
// C++ program to find the
// least non-overlapping number
// from a given set intervals
#include
using namespace std;
// function to find the smallest
// non-overlapping number
void find_missing(
vector > interval)
{
// Sort the intervals based on their
// starting value
sort(interval.begin(), interval.end());
int mx = 0;
for (int i = 0; i < (int)interval.size(); ++i) {
// check if any missing vaue exist
if (interval[i].first > mx) {
cout << mx;
return;
}
else
mx = max(mx, interval[i].second + 1);
}
// finally print the missing value
cout << mx;
}
// Driver function
int main()
{
vector > interval
= { { 0, 14 }, { 86, 108 },
{ 22, 30 }, { 5, 17 } };
find_missing(interval);
return 0;
}
Java
// Java program to find the
// least non-overlapping number
// from a given set intervals
import java.util.*;
import java.io.*;
class GFG{
static class Pair implements Comparable
{
int start,end;
Pair(int s, int e)
{
start = s;
end = e;
}
public int compareTo(Pair p)
{
return this.start - p.start;
}
}
// Function to find the smallest
// non-overlapping number
static void findMissing(ArrayList interval)
{
// Sort the intervals based on their
// starting value
Collections.sort(interval);
int mx = 0;
for(int i = 0; i < interval.size(); ++i)
{
// Check if any missing vaue exist
if (interval.get(i).start > mx)
{
System.out.println(mx);
return;
}
else
mx = Math.max(mx, interval.get(i).end + 1);
}
// Finally print the missing value
System.out.println(mx);
}
// Driver code
public static void main(String []args)
{
ArrayList interval = new ArrayList<>();
interval.add(new Pair(0, 14));
interval.add(new Pair(86, 108));
interval.add(new Pair(22, 30));
interval.add(new Pair(5, 17));
findMissing(interval);
}
}
// This code is contributed by Ganeshchowdharysadanala
Python3
# Python3 program to find the
# least non-overlapping number
# from a given set intervals
# function to find the smallest
# non-overlapping number
def find_missing(interval):
# Sort the intervals based
# on their starting value
interval.sort()
mx = 0
for i in range (len(interval)):
# Check if any missing
# vaue exist
if (interval[i][0] > mx):
print (mx)
return
else:
mx = max(mx,
interval[i][1] + 1)
# Finally print the missing value
print (mx)
# Driver code
if __name__ == "__main__":
interval = [[0, 14], [86, 108],
[22, 30], [5, 17]]
find_missing(interval);
# This code is contributed by Chitranayal
C#
// C# program to find the
// least non-overlapping number
// from a given set intervals
using System;
using System.Collections.Generic;
class GFG{
class Pair : IComparable
{
public int start,end;
public Pair(int s, int e)
{
start = s;
end = e;
}
public int CompareTo(Pair p)
{
return this.start - p.start;
}
}
// Function to find the smallest
// non-overlapping number
static void findMissing(List interval)
{
// Sort the intervals based on their
// starting value
interval.Sort();
int mx = 0;
for(int i = 0; i < interval.Count; ++i)
{
// Check if any missing vaue exist
if (interval[i].start > mx)
{
Console.WriteLine(mx);
return;
}
else
mx = Math.Max(mx, interval[i].end + 1);
}
// Finally print the missing value
Console.WriteLine(mx);
}
// Driver code
public static void Main(String []args)
{
List interval = new List();
interval.Add(new Pair(0, 14));
interval.Add(new Pair(86, 108));
interval.Add(new Pair(22, 30));
interval.Add(new Pair(5, 17));
findMissing(interval);
}
}
// This code is contributed by shikhasingrajput
输出:
18
时间复杂度: O(N * logN)
辅助空间: O(1)