给定一个整数N和两个数组增加 []和减少 [] ,这样它们只有从1 到 N 的元素。任务是为两个数组的每个元素找到达到0或N的最小步数。 步定义如下:
- 一步, increasing[]数组的所有元素都加1, decreating[]数组的所有元素都减1。
- 当一个元素变为0或N 时,不再对其执行增减操作。
例子:
Input: N = 5, increasing[] = {1, 2}, decreasing[] = {3, 4}
Output: 4
Explanation:
Step 1: increasing[] array becomes {2, 3}, decreasing[] = {2, 3}
Step 2: increasing[] array becomes {3, 4}, decreasing[] = {1, 2}
Step 3: increasing[] array becomes {4, 5}, decreasing[] = {0, 1}
Step 4: increasing[] array becomes {5, 5}, decreasing[] = {0, 0}
4 Steps are required for all elements to become either 0 or N. Hence, the output is 4.
Input: N = 7, increasing[] = {3, 5}, decreasing[] = {6}
Output: 6
方法:想法是在递增[]数组的所有元素所需的步骤之间找到最大值 和递减[]数组分别达到N和0 。以下是步骤:
- 找到数组increase[]的最小元素。
- 递增 []数组的所有元素达到N所采取的最大步长由N – min(increasing[]) 给出。
- 找到数组的最大元素下降 [] 。
- 递减[]数组的所有元素达到0所采取的最大步长由max(decresing[]) 给出。
- 因此,当所有元素变为0或N时的最小步数由max(N – min(increasing[]), max(decresing[])) 给出。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function that finds the minimum
// steps to reach either 0 or N for
// given increasing and decreasing
// arrays
void minSteps(int N, int increasing[],
int decreasing[], int m1, int m2)
{
// Initialize variable to
// find the minimum element
int mini = INT_MAX;
// Find minimum element in
// increasing[] array
for(int i = 0; i < m1; i++)
{
if (mini > increasing[i])
mini = increasing[i];
}
// Initialize variable to
// find the maximum element
int maxi = INT_MIN;
// Find maximum element in
// decreasing[] array
for(int i = 0; i < m2; i++)
{
if (maxi < decreasing[i])
maxi = decreasing[i];
}
// Find the minimum steps
int minSteps = max(maxi,
N - mini);
// Print the minimum steps
cout << minSteps << endl;
}
// Driver code
int main()
{
// Given N
int N = 7;
// Given increasing
// and decreasing array
int increasing[] = { 3, 5 };
int decreasing[] = { 6 };
// Find length of arrays
// increasing and decreasing
int m1 = sizeof(increasing) /sizeof(increasing[0]);
int m2 = sizeof(decreasing) / sizeof(decreasing[0]);
// Function call
minSteps(N, increasing, decreasing, m1, m2);
}
// This code is contributed by Manne Sree Charan
Java
// Java program for the above approach
import java.util.*;
public class GFG {
// Function that finds the minimum
// steps to reach either 0 or N for
// given increasing and decreasing
// arrays
public static void
minSteps(int N, int[] increasing,
int[] decreasing)
{
// Initialize variable to
// find the minimum element
int min = Integer.MAX_VALUE;
// Find minimum element in
// increasing[] array
for (int i : increasing) {
if (min > i)
min = i;
}
// Initialize variable to
// find the maximum element
int max = Integer.MIN_VALUE;
// Find maximum element in
// decreasing[] array
for (int i : decreasing) {
if (max < i)
max = i;
}
// Find the minimum steps
int minSteps = Math.max(max,
N - min);
// Print the minimum steps
System.out.println(minSteps);
}
// Driver Code
public static void main(String[] args)
{
// Given N
int N = 7;
// Given increasing
// and decreasing array
int increasing[] = { 3, 5 };
int decreasing[] = { 6 };
// Function call
minSteps(N, increasing, decreasing);
}
}
Python3
# Python3 program for
# the above approach
import sys
# Function that finds the minimum
# steps to reach either 0 or N for
# given increasing and decreasing
# arrays
def minSteps(N, increasing, decreasing):
# Initialize variable to
# find the minimum element
Min = sys.maxsize;
# Find minimum element in
# increasing array
for i in increasing:
if (Min > i):
Min = i;
# Initialize variable to
# find the maximum element
Max = -sys.maxsize;
# Find maximum element in
# decreasing array
for i in decreasing:
if (Max < i):
Max = i;
# Find the minimum steps
minSteps = max(Max, N - Min);
# Prthe minimum steps
print(minSteps);
# Driver Code
if __name__ == '__main__':
# Given N
N = 7;
# Given increasing
# and decreasing array
increasing = [3, 5];
decreasing = [6];
# Function call
minSteps(N, increasing, decreasing);
# This code contributed by Rajput-Ji
C#
// C# program for the above approach
using System;
class GFG{
// Function that finds the minimum
// steps to reach either 0 or N for
// given increasing and decreasing
// arrays
public static void minSteps(int N, int[] increasing,
int[] decreasing)
{
// Initialize variable to
// find the minimum element
int min = int.MaxValue;
// Find minimum element in
// increasing[] array
foreach(int i in increasing)
{
if (min > i)
min = i;
}
// Initialize variable to
// find the maximum element
int max = int.MinValue;
// Find maximum element in
// decreasing[] array
foreach(int i in decreasing)
{
if (max < i)
max = i;
}
// Find the minimum steps
int minSteps = Math.Max(max,
N - min);
// Print the minimum steps
Console.WriteLine(minSteps);
}
// Driver Code
public static void Main(String[] args)
{
// Given N
int N = 7;
// Given increasing
// and decreasing array
int []increasing = { 3, 5 };
int []decreasing = { 6 };
// Function call
minSteps(N, increasing, decreasing);
}
}
// This code is contributed by Amit Katiyar
Javascript
6
时间复杂度: O(N)
辅助空间: O(1)
如果您想与行业专家一起参加直播课程,请参阅Geeks Classes Live