给定一个由N 个整数组成的数组arr[]和一个由K个类型为{L, R} 的查询组成的数组Q[][2] ,每个查询的任务是检查子数组{arr[L], ..数组的arr[R]}是否非递减。如果发现是真的,则打印“是” 。否则,打印“否”。
例子:
Input: arr[] = {1, 7, 3, 4, 9}, K = 2, Q[][] = {{1, 2}, {2, 4}}
Output:
Yes
No
Explanation:
Query 1: The subarray in the range [1, 2] is {1, 7} which is non-decreasing. Therefore, print “Yes”.
Query 2: The subarray in the range [2, 4] is {7, 3, 4, 9} which is not non-decreasing. Therefore, print “No”.
Input: arr[] = {3, 5, 7, 1, 8, 2}, K = 3, Q[][] = {{1, 3}, {2, 5}, {4, 6}}
Output:
Yes
No
No
朴素的方法:最简单的方法是在每个查询的索引范围[L, R]上遍历数组,并检查子数组是否按升序排序。如果发现是真的,则打印“是” 。否则,打印“否”。
时间复杂度: O(N * Q)
辅助空间: O(1)
高效的方法:可以通过预先计算范围[1, i]中满足arr[i] > arr[i + 1]的相邻元素的数量来优化上述方法,这导致对范围内此类索引的数量进行恒定时间计算[L, R – 1] 。请按照以下步骤解决问题:
- 初始化一个数组,比如pre[] ,以存储从起始索引开始的索引计数,相邻元素按递增顺序排列。
- 迭代范围[1, N – 1]并分配pre[i] = pre[i – 1]然后将pre[i]增加1 ,如果arr[i – 1] > arr[i] 。
- 遍历数组Q[][]并对每个查询{L, R} ,如果pre[R – 1] – pre[L – 1]为0 ,则打印“Yes” 。否则,打印“否”。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to perform queries to check if
// subarrays over a given range of indices
// is non-decreasing or not
void checkSorted(int arr[], int N,
vector >& Q)
{
// Stores count of indices up to i
// such that arr[i] > arr[i + 1]
int pre[N] = { 0 };
// Traverse the array
for (int i = 1; i < N; i++) {
// Update pre[i]
pre[i] = pre[i - 1]
+ (arr[i - 1] > arr[i]);
}
// Traverse the array Q[][]
for (int i = 0; i < Q.size(); i++) {
int l = Q[i][0];
int r = Q[i][1] - 1;
// If pre[r] - pre[l-1] exceeds 0
if (pre[r] - pre[l - 1] == 0)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
}
// Driver Code
int main()
{
int arr[] = { 1, 7, 3, 4, 9 };
vector > Q = { { 1, 2 },
{ 2, 4 } };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
checkSorted(arr, N, Q);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG
{
// Function to perform queries to check if
// subarrays over a given range of indices
// is non-decreasing or not
static void checkSorted(int[] arr, int N, int[][] Q)
{
// Stores count of indices up to i
// such that arr[i] > arr[i + 1]
int[] pre = new int[N];
// Traverse the array
for (int i = 1; i < N; i++)
{
// Update pre[i]
if((arr[i - 1] > arr[i]))
pre[i] = pre[i - 1] + 1;
else
pre[i] = pre[i - 1];
}
// Traverse the array Q[][]
for (int i = 0; i < Q.length; i++)
{
int l = Q[i][0];
int r = Q[i][1] - 1;
// If pre[r] - pre[l-1] exceeds 0
if (pre[r] - pre[l - 1] == 0)
System.out.println("Yes");
else
System.out.println("No");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 7, 3, 4, 9 };
int Q[][] = { { 1, 2 }, { 2, 4 } };
int N = arr.length;
// Function Call
checkSorted(arr, N, Q);
}
}
// This code is contributed by Dharanendra L V.
Python3
# Python3 program for the above approach
# Function to perform queries to check if
# subarrays over a given range of indices
# is non-decreasing or not
def checkSorted(arr, N, Q):
# Stores count of indices up to i
# such that arr[i] > arr[i + 1]
pre = [0]*(N)
# Traverse the array
for i in range(1, N):
# Update pre[i]
pre[i] = pre[i - 1] + (arr[i - 1] > arr[i])
# Traverse the array Q[][]
for i in range(len(Q)):
l = Q[i][0]
r = Q[i][1] - 1
# If pre[r] - pre[l-1] exceeds 0
if (pre[r] - pre[l - 1] == 0):
print("Yes")
else:
print("No")
# Driver Code
if __name__ == '__main__':
arr =[1, 7, 3, 4, 9]
Q = [ [ 1, 2 ],[ 2, 4 ] ]
N = len(arr)
# Function Call
checkSorted(arr, N, Q)
# This code is contributed by mohit kumar 29.
C#
// C# program for the above approach
using System;
public class GFG{
// Function to perform queries to check if
// subarrays over a given range of indices
// is non-decreasing or not
static void checkSorted(int[] arr, int N, int[,] Q)
{
// Stores count of indices up to i
// such that arr[i] > arr[i + 1]
int[] pre = new int[N];
// Traverse the array
for (int i = 1; i < N; i++)
{
// Update pre[i]
if((arr[i - 1] > arr[i]))
{
pre[i] = pre[i - 1] + 1;
}
else
{ pre[i] = pre[i - 1];}
}
// Traverse the array Q[][]
for (int i = 0; i < Q.GetLength(0); i++)
{
int l = Q[i,0];
int r = Q[i,1] - 1;
// If pre[r] - pre[l-1] exceeds 0
if (pre[r] - pre[l - 1] == 0)
{ Console.WriteLine("Yes");}
else
{Console.WriteLine("No");}
}
}
// Driver Code
static public void Main (){
int[] arr = { 1, 7, 3, 4, 9 };
int[,] Q = { { 1, 2 }, { 2, 4 } };
int N = arr.Length;
// Function Call
checkSorted(arr, N, Q);
}
}
// This code is contributed by avanitrachhadiya2155
Yes
No
时间复杂度: O(N)
辅助空间: O(1)
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