执行查询以将 K 添加到索引范围 [L, R] 后数组中存在的最大元素

给定一个由N个整数组成的数组arr[] （最初设置为 0 ）和一个数组Q[] ，由{l, r, k}形式的查询组成，每个查询的任务是将K添加到索引l到r （包括两者）。执行所有查询后，返回数组中存在的最大元素。

例子：

Input: N=10, q[] = {{1, 5, 3}, {4, 8, 7}, {6, 9, 1}}
Output: 10
Explanation: 
Initially the array is → [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Query1 {1, 5, 3} results in [3, 3, 3, 3, 3, 0, 0, 0, 0, 0]
Query2 {4, 8, 7} results in [3, 3, 3, 10, 10, 7, 7, 7, 0, 0]
Query2 {6, 9, 1} results in [3, 3, 3, 10, 10, 8, 8, 8, 1, 0]
Maximum value in the updated array = 10

方法：按照以下步骤解决问题。

• 遍历查询向量和每个查询{l, r, k}
  • 将k添加到a[l]并从a[r+1]中减去 k
• 初始化变量x = 0来存储运行总和， m = INT_MIN来存储最大值
• 遍历数组，向x添加元素，并更新m。
• 打印最大值m

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
// Function to find the max sum
// after processing q queries
int max_sum(int a[],
            vector, int> > v,
            int q, int n)
{
    // Store the cumulative sum
    int x = 0;
    // Store the maximum sum
    int m = INT_MIN;
 
    // Iterate over the range 0 to q
    for (int i = 0; i < q; i++) {
 
        // Variables to extract
        // values from vector
        int p, q, k;
 
        p = v[i].first.first;
        q = v[i].first.second;
        k = v[i].second;
        a[p] += k;
 
        if (q + 1 <= n)
 
            a[q + 1] -= k;
    }
 
    // Iterate over the range [1, n]
    for (int i = 1; i <= n; i++)
 
    {
        // Calculate cumulative sum
        x += a[i];
 
        // Calculate maximum sum
        m = max(m, x);
    }
    // Return the maximum sum after q queries
    return m;
}
 
// Driver code
int main()
{
 
    // Stores the size of array
    // and number of queries
    int n = 10, q = 3;
 
    // Stores the sum
    int a[n + 5] = { 0 };
 
    // Storing input queries
    vector, int> > v(q);
    v[0].first.first = 1;
    v[0].first.second = 5;
    v[0].second = 3;
    v[1].first.first = 4;
    v[1].first.second = 8;
    v[1].second = 7;
    v[2].first.first = 6;
    v[2].first.second = 9;
    v[2].second = 1;
 
    // Function call to find the maximum sum
    cout << max_sum(a, v, q, n);
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
     
// Function to find the max sum
// after processing q queries
static int max_sum(int a[],
                   ArrayList> v,
                   int q, int n)
{
     
    // Store the cumulative sum
    int x = 0;
     
    // Store the maximum sum
    int m = Integer.MIN_VALUE;
 
    // Iterate over the range 0 to q
    for(int i = 0; i < q; i++)
    {
         
        // Variables to extract
        // values from vector
        int p, qq, k;
 
        p = v.get(i).get(0);
        qq = v.get(i).get(1);
        k = v.get(i).get(2);
        a[p] += k;
 
        if (qq + 1 <= n)
            a[qq + 1] -= k;
    }
 
    // Iterate over the range [1, n]
    for(int i = 1; i <= n; i++)
    {
         
        // Calculate cumulative sum
        x += a[i];
 
        // Calculate maximum sum
        m = Math.max(m, x);
    }
     
    // Return the maximum sum after q queries
    return m;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Stores the size of array
    // and number of queries
    int n = 10, q = 3;
     
    // Stores the sum
    int[] a = new int[n + 5];
     
    // Storing input queries
    ArrayList> v= new ArrayList<>();
     
    for(int i = 0; i < q; i++)
        v.add(new ArrayList<>());
     
    v.get(0).add(1);
    v.get(0).add(5);
    v.get(0).add(3);
    v.get(1).add(4);
    v.get(1).add(8);
    v.get(1).add(7);
    v.get(2).add(6);
    v.get(2).add(9);
    v.get(2).add(1);
     
    // Function call to find the maximum sum
    System.out.println(max_sum(a, v, q, n));
}
}
 
// This code is contributed by offbeat

Python3

# Python program for the above approach
 
# Function to find the max sum
# after processing q queries
def max_sum(a, v, q, n):
   
    # Store the cumulative sum
    x = 0;
     
    # Store the maximum sum
    m = -10**9;
 
    # Iterate over the range 0 to q
    for i in range(q):
 
        # Variables to extract
        # values from vector
        p = v[i][0][0];
        q = v[i][0][1];
        k = v[i][1];
        a[p] += k;
 
        if (q + 1 <= n):
            a[q + 1] -= k;
 
    # Iterate over the range [1, n]
    for i in range(1, n + 1):
        # Calculate cumulative sum
        x += a[i];
 
        # Calculate maximum sum
        m = max(m, x);
 
    # Return the maximum sum after q queries
    return m;
 
# Driver code
 
# Stores the size of array
# and number of queries
n = 10
q = 3;
 
# Stores the sum
a = [0] * (n + 5);
 
# Storing input queries
v = [[[0 for i in range(2)] for x in range(2)] for z in range(q)]
v[0][0][0] = 1;
v[0][0][1] = 5;
v[0][1] = 3;
v[1][0][0] = 4;
v[1][0][1] = 8;
v[1][1] = 7;
v[2][0][0] = 6;
v[2][0][1] = 9;
v[2][1] = 1;
 
# Function call to find the maximum sum
print(max_sum(a, v, q, n));
 
# This code is contributed by _saurabh_jaiswal

Javascript

// Javascript program for the above approach
 
 
// Function to find the max sum
// after processing q queries
function max_sum(a, v, q, n) {
    // Store the cumulative sum
    let x = 0;
    // Store the maximum sum
    let m = Number.MIN_SAFE_INTEGER;
 
    // Iterate over the range 0 to q
    for (let i = 0; i < q; i++) {
 
        // Variables to extract
        // values from vector
        let p, q, k;
 
        p = v[i][0][0];
        q = v[i][0][1];
        k = v[i][1];
        a[p] += k;
 
        if (q + 1 <= n)
 
            a[q + 1] -= k;
    }
 
    // Iterate over the range [1, n]
    for (let i = 1; i <= n; i++) {
        // Calculate cumulative sum
        x += a[i];
 
        // Calculate maximum sum
        m = Math.max(m, x);
    }
    // Return the maximum sum after q queries
    return m;
}
 
// Driver code
 
 
// Stores the size of array
// and number of queries
let n = 10, q = 3;
 
// Stores the sum
let a = new Array(n + 5).fill(0);
 
// Storing input queries
let v = new Array(q).fill(0).map(() => new Array(2).fill(0).map(() => new Array(2).fill(0)));
v[0][0][0] = 1;
v[0][0][1] = 5;
v[0][1] = 3;
v[1][0][0] = 4;
v[1][0][1] = 8;
v[1][1] = 7;
v[2][0][0] = 6;
v[2][0][1] = 9;
v[2][1] = 1;
 
// Function call to find the maximum sum
document.write(max_sum(a, v, q, n));
 
// This code is contributed by gfgking.

输出：

10

时间复杂度： O(N+K) 其中 N 是数组的大小，K 是查询的数量
空间复杂度： O(1)