将数组的所有元素减少为零所需的最少步骤

本文概述

C ++
Java
Python3
C#

给定一个数组arr []对正整数, 任务是找到最小的步骤, 以减少所有元素为0。一步, -1可以同时添加到数组的所有非零元素中。
例子：

输入：arr [] = {1, 5, 6}
输出：6
操作1：arr [] = {0, 4, 5}
操作2：arr [] = {0, 3, 4}
操作3：arr [] = {0, 2, 3}
运算4：arr [] = {0, 1, 2}
运算5：arr [] = {0, 0, 1}
运算6：arr [] = {0, 0, 0}
输入：arr [] = {1, 1}
输出：1

简单的方法：一种简单的方法是先对数组进行排序, 然后从最小元素开始, 计算将其减少到0所需的步骤数。然后, 从下一个数组元素开始减少此计数, 因为所有元素将同时更新。
高效的方法：可以看到, 最小步数将始终等于数组中的最大元素。
下面是上述方法的实现：
C ++

//C++ implementation of the approach #include < bits/stdc++.h> using namespace std; //Function to return the minimum steps //required to reduce all the elements to 0 int minSteps( int arr[], int n) {//Maximum element from the array int maxVal = *max_element(arr, arr + n); return maxVal; }//Driver code int main() { int arr[] = { 1, 2, 4 }; int n = sizeof (arr) /sizeof ( int ); cout < < minSteps(arr, n); return 0; }

Java

//Java implementation of the approach class GFG {//method to get maximum number from array elements static int getMax( int inputArray []) { int maxValue = https://www.lsbin.com/inputArray[ 0 ]; for ( int i = 1 ; i < inputArray.length; i++) { if (inputArray[i]> maxValue) { maxValue = inputArray[i]; } } return maxValue; }//Function to return the minimum steps //required to reduce all the elements to 0 static int minSteps( int arr[], int n) { //Maximum element from the array int maxVal = getMax(arr); return maxVal; } //Driver code public static void main (String[] args) { int arr[] = { 1 , 2 , 4 }; int n = arr.length; System.out.println(minSteps(arr, n)); } }//This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach# Function to return the minimum steps # required to reduce all the elements to 0 def minSteps(arr, n):# Maximum element from the array maxVal = max (arr) return maxVal# Driver code arr = [ 1 , 2 , 4 ] n = len (arr)print (minSteps(arr, n))# This code is contributed by Mohit Kumar

//C# implementation of the approach using System; class GFG {//method to get maximum number from array elements static int getMax( int []inputArray) { int maxValue = https://www.lsbin.com/inputArray[0]; for ( int i = 1; i < inputArray.Length; i++) { if (inputArray[i]> maxValue) { maxValue = inputArray[i]; } } return maxValue; }//Function to return the minimum steps //required to reduce all the elements to 0 static int minSteps( int []arr, int n) { //Maximum element from the array int maxVal = getMax(arr); return maxVal; } //Driver code public static void Main(String []args) { int []arr = { 1, 2, 4 }; int n = arr.Length; Console.WriteLine(minSteps(arr, n)); } }//This code is contributed by Arnab Kundu

【将数组的所有元素减少为零所需的最少步骤】输出如下：