在执行查询以将(I–L+1)添加到范围[L,R]
中存在的每个元素后,打印修改后的数组
原文:https://www . geesforgeks . org/print-modified-array-执行查询后-add-I-l-1-to-每个元素-范围内存在-l-r/
给定由NT6】0s(1 基索引)和另一个数组查询【】组成的数组arr【】,表格的每一行 {L,R} ,每个查询 (L,R) 的任务是在范围上添加一个值(I–L+1)****
示例:
输入: arr[] = {0,0,0},query[][] = {{1,3},{2,3}} 输出: 1 3 5 解释:最初数组为{0,0,0}。 查询 1: 涉及的指标范围:[1,3]。范围内每个索引 I 的值(I–1+1)为{1,2,3}。添加这些值会将数组修改为{1,2,3}。 查询 2: 涉及的指标范围:[2,3]。范围内每个索引 I 的值(I–2+1)为{0,1,2}。添加这些值会将数组修改为{1,3,5}。 因此,修改后的数组为{1,3,5}。
输入: arr[] = {0,0,0,0,0,0,0},查询[][] = {{1,7},{3,6},{4,5}} 输出: 1 2 4 7 10 10 7
天真方法:解决给定问题的最简单方法是遍历给定数组的范围【L,R】并将值(I–L+1)添加到每个查询的范围内的每个元素。完成所有查询后,打印修改后的数组得到 arr[] 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector<vector<int> > queries,
int N)
{
// Initialize an array a[]
vector<int> a(N + 1, 0);
// Stores the size of the array
int n = N + 1;
int q = queries.size();
// Traverse the queries
for (int i = 0; i < q; i++) {
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment each index from L to
// R in a[] by (j - l + 1)
for (int j = l; j <= r; j++) {
a[j] += (j - l + 1);
}
}
// Print the modified array
for (int i = 1; i <= N; i++) {
cout << a[i] << " ";
}
}
// Driver Code
int main()
{
int N = 7;
vector<vector<int> > queries
= { { 1, 7 }, { 3, 6 }, { 4, 5 } };
// Function Call
updateQuery(queries, N);
return 0;
}
Java 语言(一种计算机语言,尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to perform the given queries
// in the given empty array of size N
static void updateQuery(int [][]queries, int N)
{
// Initialize an array a[]
ArrayList<Integer> a = new ArrayList<Integer>();
for(int i = 0; i < N + 1; i++)
a.add(0);
// Stores the size of the array
int q = 3;
// Traverse the queries
for(int i = 0; i < q; i++)
{
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment each index from L to
// R in a[] by (j - l + 1)
for(int j = l; j <= r; j++)
{
a.set(j, a.get(j)+(j - l + 1));
}
}
// Print the modified array
for(int i = 1; i < a.size(); i++)
{
System.out.print(a.get(i) + " ");
}
}
// Driver code
public static void main(String[] args)
{
int N = 7;
int[][] queries = { { 1, 7 },
{ 3, 6 },
{ 4, 5 } };
// Function Call
updateQuery(queries, N);
}
}
// This code is contributed by offbeat
Python 3
# Python 3 program for the above approach
# Function to perform the given queries
# in the given empty array of size N
def updateQuery(queries, N):
# Initialize an array a[]
a = [0 for i in range(N + 1)]
# Stores the size of the array
n = N + 1
q = len(queries)
# Traverse the queries
for i in range(q):
# Starting index
l = queries[i][0]
# Ending index
r = queries[i][1]
# Increment each index from L to
# R in a[] by (j - l + 1)
for j in range(l, r + 1, 1):
a[j] += (j - l + 1)
# Print the modified array
for i in range(1, N + 1, 1):
print(a[i], end = " ")
# Driver Code
if __name__ == '__main__':
N = 7
queries = [[1, 7],[3, 6],[4, 5]]
# Function Call
updateQuery(queries, N)
# This code is contributed by ipg2016107.
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to perform the given queries
// in the given empty array of size N
static void updateQuery(int [,]queries, int N)
{
// Initialize an array a[]
List<int> a = new List<int>();
for(int i = 0; i < N + 1; i++)
a.Add(0);
// Stores the size of the array
int q = 3;
// Traverse the queries
for(int i = 0; i < q; i++)
{
// Starting index
int l = queries[i, 0];
// Ending index
int r = queries[i, 1];
// Increment each index from L to
// R in a[] by (j - l + 1)
for(int j = l; j <= r; j++)
{
a[j] += (j - l + 1);
}
}
// Print the modified array
for(int i = 1; i < a.Count; i++)
{
Console.Write(a[i] + " ");
}
}
// Driver Code
public static void Main()
{
int N = 7;
int[,] queries = new int[3, 2] { { 1, 7 },
{ 3, 6 },
{ 4, 5 } };
// Function Call
updateQuery(queries, N);
}
}
// This code is contributed by SURENDRA_GANGWAR
Output:
1 2 4 7 10 10 7
时间复杂度: O(NQ)* 辅助空间: O(1)
高效方法:上述方法可以通过使用前缀和进行优化。按照以下步骤解决此问题:
- 初始化一个数组ans【】,所有元素为 0 ,以存储当前索引受查询影响的次数。
- 初始化一个数组 res[] ,所有元素为 0 ,存储某个查询范围结束后要删除的值。
- 遍历给定的查询数组查询[] ,并执行以下步骤:
- 将 1 加到 ans【查询[I][0]】上,从 ans【查询[I][1]+1】中减去 1 。
- 从 res【查询[I][1]+1】中减去(查询[I][1]–查询[i][0] + 1) 。
- 找到数组和的前缀和。
- 遍历数组 res[] 并将每个元素 res[i] 更新为RES[I]+RES[I–1]+ans[I]。
- 完成上述步骤后,在执行给定查询后,将数组 res[] 打印为修改后的数组。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector<vector<int> > queries,
int N)
{
// Stores the resultant array
// and the difference array
vector<int> ans(N + 2, 0),
res(N + 2, 0);
int q = queries.size();
// Traverse the given queries
for (int i = 0; i < q; i++) {
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment l-th index by 1
ans[l]++;
// Decrease r-th index by 1
ans[r + 1]--;
// Decrease (r + 1)th index by
// the length of each query
res[r + 1] -= (r - l + 1);
}
// Find the prefix sum of ans[]
for (int i = 1; i <= N; i++)
ans[i] += ans[i - 1];
// Find the final array
for (int i = 1; i <= N; i++)
res[i] += res[i - 1] + ans[i];
// Printing the modified array
for (int i = 1; i <= N; i++) {
cout << res[i] << " ";
}
cout << "\n";
}
// Driver Code
int main()
{
int N = 7;
vector<vector<int> > queries
= { { 1, 7 }, { 3, 6 }, { 4, 5 } };
updateQuery(queries, N);
return 0;
}
Output:
1 2 4 7 10 10 7
时间复杂度:O(N) T5辅助空间:** O(N)
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