给定n个塔的高度和一个值k。我们需要将每个塔的高度增加或减少k(仅一次),其中k>0。任务是在修改后将最长和最短塔的高度差降至最低,并输出该差。 例如:
null
Input : arr[] = {1, 15, 10}, k = 6Output : Maximum difference is 5.Explanation : We change 1 to 7, 15 to 9 and 10 to 4. Maximum difference is 5(between 4 and 9). We can't get a lowerdifference.Input : arr[] = {1, 5, 15, 10} k = 3 Output : Maximum difference is 8arr[] = {4, 8, 12, 7}Input : arr[] = {4, 6} k = 10Output : Maximum difference is 2arr[] = {14, 16} OR {-6, -4}Input : arr[] = {6, 10} k = 3Output : Maximum difference is 2arr[] = {9, 7} Input : arr[] = {1, 10, 14, 14, 14, 15} k = 6 Output: Maximum difference is 5arr[] = {7, 4, 8, 8, 8, 9} Input : arr[] = {1, 2, 3} k = 2 Output: Maximum difference is 2arr[] = {3, 4, 5}
资料来源: Adobe面试体验|第24集(MTS校园)
首先,我们尝试对阵列进行排序,并使每个塔的高度最大。我们通过将所有塔楼的高度向右降低k,并将所有塔楼的高度向左增加k来实现这一点。你试图增加高度的塔也可能没有最大高度。因此,我们只需将其与右侧的最后一个元素(a[n]-k)进行比较,来检查它是否具有最大高度。因为阵列是排序的,如果塔的高度大于a[n]-k,那么它就是可用的最高塔。类似的推理也可用于寻找最短的塔。
注意:我们不需要考虑在哪里[i]<k,因为塔的高度不能是负的,所以我们必须忽略这种情况。
时间复杂性: O(nlogn)
空间复杂性: O(n)
C++
#include <bits/stdc++.h> using namespace std; // User function Template int getMinDiff( int arr[], int n, int k) { sort(arr, arr + n); int ans = arr[n - 1] - arr[0]; // Maximum possible height difference int tempmin, tempmax; tempmin = arr[0]; tempmax = arr[n - 1]; for ( int i = 1; i < n; i++) { tempmin= min(arr[0] + k,arr[i] - k); // Minimum element when we // add k to whole array tempmax = max(arr[i - 1] + k, arr[n - 1] - k); // Maximum element when we // subtract k from whole array ans = min(ans, tempmax - tempmin); } return ans; } // Driver Code Starts int main() { int k = 6, n = 6; int arr[n] = { 7, 4, 8, 8, 8, 9 }; int ans = getMinDiff(arr, n, k); cout << ans; } |
JAVA
/*package whatever //do not write package name here */ // { Driver Code Starts // Initial Template for Java import java.io.*; import java.util.*; public class Main { public static void main(String[] args) { int [] arr = { 7 , 4 , 8 , 8 , 8 , 9 }; int k = 6 ; int ans = getMinDiff(arr, arr.length, k); System.out.println(ans); } // } Driver Code Ends // User function Template for Java public static int getMinDiff( int [] arr, int n, int k) { Arrays.sort(arr); int ans = (arr[n - 1 ] + k)- (arr[ 0 ] + k); // Maximum possible height difference int tempmax = arr[n - 1 ] - k; // Maximum element when we // subtract k from whole array int tempmin = arr[ 0 ] + k; // Minimum element when we // add k to whole array int max, min; for ( int i = 0 ; i < n - 1 ; i++) { if (tempmax > (arr[i] + k)) { max = tempmax; } else { max = arr[i] + k; } if (tempmin < (arr[i + 1 ] - k)) { min = tempmin; } else { min = arr[i + 1 ] - k; } if (ans > (max - min)) { ans = max - min; } } return ans; } } |
Python3
# User function Template def getMinDiff(arr, n, k): arr.sort() ans = arr[n - 1 ] - arr[ 0 ] # Maximum possible height difference tempmin = arr[ 0 ] tempmax = arr[n - 1 ] for i in range ( 1 , n): tempmin = min (arr[ 0 ] + k, arr[i] - k) # Minimum element when we # add k to whole array # Maximum element when we tempmax = max (arr[i - 1 ] + k, arr[n - 1 ] - k) # subtract k from whole array ans = min (ans, tempmax - tempmin) return ans # Driver Code Starts k = 6 n = 6 arr = [ 7 , 4 , 8 , 8 , 8 , 9 ] ans = getMinDiff(arr, n, k) print (ans) # This code is contributed by ninja_hattori. |
Javascript
<script> // User function Template function getMinDiff(arr,n,k) { arr.sort((a,b) => (a-b)) let ans = arr[n - 1] - arr[0]; // Maximum possible height difference let tempmin, tempmax; tempmin = arr[0]; tempmax = arr[n - 1]; for (let i = 1; i < n; i++) { tempmin= Math.min(arr[0] + k,arr[i] - k); // Minimum element when we // add k to whole array tempmax = Math.max(arr[i - 1] + k, arr[n - 1] - k); // Maximum element when we // subtract k from whole array ans = Math.min(ans, tempmax - tempmin); } return ans; } // Driver Code Starts let k = 6, n = 6; let arr = [ 7, 4, 8, 8, 8, 9 ]; let ans = getMinDiff(arr, n, k); document.write(ans, "</br>" ); //This code is contributed by shinjanpatra. </script> |
输出
5
© 版权声明
文章版权归作者所有,未经允许请勿转载。
THE END