执行加减命令后打印修改后的数组
原文: https://www.geeksforgeeks.org/print-modified-array-executing-commands-addition-subtraction/
给定大小为n
的数组和给定的大小为m
的命令集。 每个命令由四个整数q, l, r, k
组成。 这些命令具有以下类型:
-
如果
q = 0
,则将k
添加到范围a
至b
的所有整数中(1 <= a <= b <= n
)。 -
如果
q = 1
,则将a
到b
范围内的所有整数减去k
(1 <= a <= b <= n
)。
注意:最初,所有数组元素都设置为 0,而数组索引从 1 开始。
Input : n = 5
commands[] = {{0 1 3 2}, {1 3 5 3},
{0 2 5 1}}
Output : 0 2 -1 -1 -3
Explanation
First command: Add 2 from index 1 to 3
>= 2 2 2 0 0
Second command: Subtract 3 from index 3 to 5
>= 2 2 -1 -3 -3
Third command: Add 1 from index 2 to 5
>= 2 3 0 -2 -2
简单方法是通过从左索引(l
)到右索引(r
)进行迭代来执行每个命令,并根据给定命令更新每个索引。 该方法的时间复杂度为O(n * m)
。
更好的方法是使用分域树(BIT)或段树。 但这只会最优化log(n)
时间,即整体复杂度将变为O(m * log(n))
。
高效方法是使用简单的数学方法。 由于所有命令都可以离线处理,因此我们可以将所有更新存储在临时数组中,然后最后执行它。
-
对于命令 0,在第
l
个第r + 1
个索引元素中添加+k
和-k
。 -
对于命令 1,在第
l
个第r + 1
个索引元素中添加+k
和-k
。
之后,对每个索引i
的所有元素求和,从第一个索引开始,即a[i]
将包含索引从 1 到i
的所有元素的和。 这可以通过动态规划轻松实现。
C++
// C++ program to find modified array
// after executing m commands/queries
#include<bits/stdc++.h>
using namespace std;
// Update function for every command
void updateQuery(int arr[], int n, int q, int l,
int r, int k)
{
// If q == 0, add 'k' and '-k'
// to 'l-1' and 'r' index
if (q == 0){
arr[l-1] += k;
arr[r] += -k;
}
// If q == 1, add '-k' and 'k'
// to 'l-1' and 'r' index
else{
arr[l-1] += -k;
arr[r] += k;
}
return;
}
// Function to generate the final
// array after executing all
// commands
void generateArray(int arr[], int n)
{
// Generate final array with the
// help of DP concept
for (int i = 1; i < n; ++i)
arr[i] += arr[i-1];
return;
}
// Driver program
int main()
{
int n = 5;
int arr[n+1];
//Set all array elements to '0'
memset(arr, 0, sizeof(arr));
int q = 0, l = 1, r = 3, k = 2;
updateQuery(arr, n, q, l, r, k);
q = 1 , l = 3, r = 5, k = 3;
updateQuery(arr, n, q, l, r, k);
q = 0 , l = 2, r = 5, k = 1;
updateQuery(arr, n, q, l, r, k);
// Generate final array
generateArray(arr, n);
// Printing the final modified array
for (int i = 0; i < n; ++i)
cout << arr[i] << " ";
return 0;
}
Java
// Java program to find modified array
// after executing m commands/queries
import java.util.Arrays;
class GFG {
// Update function for every command
static void updateQuery(int arr[], int n,
int q, int l, int r, int k)
{
// If q == 0, add 'k' and '-k'
// to 'l-1' and 'r' index
if (q == 0){
arr[l-1] += k;
arr[r] += -k;
}
// If q == 1, add '-k' and 'k'
// to 'l-1' and 'r' index
else{
arr[l-1] += -k;
arr[r] += k;
}
return;
}
// Function to generate the final
// array after executing all
// commands
static void generateArray(int arr[], int n)
{
// Generate final array with the
// help of DP concept
for (int i = 1; i < n; ++i)
arr[i] += arr[i-1];
}
//driver code
public static void main(String arg[])
{
int n = 5;
int arr[] = new int[n+1];
//Set all array elements to '0'
Arrays.fill(arr, 0);
int q = 0, l = 1, r = 3, k = 2;
updateQuery(arr, n, q, l, r, k);
q = 1 ; l = 3; r = 5; k = 3;
updateQuery(arr, n, q, l, r, k);
q = 0 ; l = 2; r = 5; k = 1;
updateQuery(arr, n, q, l, r, k);
// Generate final array
generateArray(arr, n);
// Printing the final modified array
for (int i = 0; i < n; ++i)
System.out.print(arr[i]+" ");
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to find modified array
# after executing m commands/queries
# Update function for every command
def updateQuery(arr, n, q, l, r, k):
# If q == 0, add 'k' and '-k'
# to 'l-1' and 'r' index
if (q == 0):
arr[l - 1] += k
arr[r] += -k
# If q == 1, add '-k' and 'k'
# to 'l-1' and 'r' index
else:
arr[l - 1] += -k
arr[r] += k
return
# Function to generate the final
# array after executing all commands
def generateArray(arr, n):
# Generate final array with the
# help of DP concept
for i in range(1, n):
arr[i] += arr[i - 1]
return
# Driver Code
n = 5
arr = [0 for i in range(n + 1)]
# Set all array elements to '0'
q = 0; l = 1; r = 3; k = 2
updateQuery(arr, n, q, l, r, k)
q, l, r, k = 1, 3, 5, 3
updateQuery(arr, n, q, l, r, k);
q, l, r, k = 0, 2, 5, 1
updateQuery(arr, n, q, l, r, k)
# Generate final array
generateArray(arr, n)
# Printing the final modified array
for i in range(n):
print(arr[i], end = " ")
# This code is contributed by Anant Agarwal.
C
// Program to find modified
// array after executing
// m commands/queries
using System;
class GFG {
// Update function for every command
static void updateQuery(int[] arr, int n, int q,
int l, int r, int k)
{
// If q == 0, add 'k' and '-k'
// to 'l-1' and 'r' index
if (q == 0) {
arr[l - 1] += k;
arr[r] += -k;
}
// If q == 1, add '-k' and 'k'
// to 'l-1' and 'r' index
else {
arr[l - 1] += -k;
arr[r] += k;
}
return;
}
// Function to generate final
// array after executing all
// commands
static void generateArray(int[] arr, int n)
{
// Generate final array with
// the help of DP concept
for (int i = 1; i < n; ++i)
arr[i] += arr[i - 1];
}
// Driver code
public static void Main()
{
int n = 5;
int[] arr = new int[n + 1];
// Set all array elements to '0'
for (int i = 0; i < arr.Length; i++)
arr[i] = 0;
int q = 0, l = 1, r = 3, k = 2;
updateQuery(arr, n, q, l, r, k);
q = 1;
l = 3;
r = 5;
k = 3;
updateQuery(arr, n, q, l, r, k);
q = 0;
l = 2;
r = 5;
k = 1;
updateQuery(arr, n, q, l, r, k);
// Generate final array
generateArray(arr, n);
// Printing the final modified array
for (int i = 0; i < n; ++i)
Console.Write(arr[i] + " ");
}
}
// This code is contributed by Anant Agarwal.
输出:
Output: 2 3 0 -2 -2
时间复杂度:O(Max(m, n))
。
辅助空间:O(n)
。
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