如果 arr[i]> 2*arr[i-1]，则检查总和等于 k 的子序列是否存在

给定一个排序的正整数数组，其中arr[i] > 2*arr[i-1] ，检查是否存在总和等于 k 的子序列。
例子：

Input : arr[]={ 1, 3, 7, 15, 31}, K=18 
Output :True 
A[1] + A[3] = 3 + 15 = 18 
We found a subsequence whose sum is 18 
Input :arr[]={ 1, 3, 7, 15, 31}, K=20 
Output :False 
No subsequence can be found with sum 20 
 

朴素的解决方案：基本的解决方案是检查所有 2^n 种可能的组合，并检查是否存在总和等于 K 的子序列。此过程不适用于更高的 N 值，N>20。
时间复杂度：O(2^N)
有效的解决方案：我们得到arr[i] >2*arr[i-1]所以我们可以说arr[i] > ( arr[i-1] + arr[i-2] + …+ arr[2] + arr[1] + arr[0] ) 。
让我们假设 arr[i] <= K ( arr[i-1] + arr[i-2] + …+ arr[2] + arr[1] + arr[0] ) )，所以我们必须包括arr[i] 在集合中。因此，我们必须以降序遍历数组，当我们找到 arr[i]<=k 时，我们将 arr[i] 包含在集合中并从 K 中减去 arr[i] 并继续循环，直到 K 的值是等于零。
如果 K 的值为零，则存在子序列，否则不存在。
以下是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to check whether sum of any set
// of the array element is equal
// to k or not
bool CheckForSequence(int arr[], int n, int k)
{
    // Traverse the array from end
    // to start
    for (int i = n - 1; i >= 0; i--) {
        // if k is greater than
        // arr[i] then subtract
        // it from k
        if (k >= arr[i])
            k -= arr[i];
    }
 
    // If there is any subsequence
    // whose sum is equal to k
    if (k != 0)
        return false;
    else
        return true;
}
 
// Driver code
int main()
{
    int A[] = { 1, 3, 7, 15, 31 };
    int n = sizeof(A) / sizeof(int);
    cout << (CheckForSequence(A, n, 18)
                 ? "True": "False") << endl;
    return 0;
}

Java

// Java implementation of above approach
import java.io.*;
 
class GFG
{
     
// Function to check whether
// sum of any set of the array element
// is equal to k or not
static boolean CheckForSequence(int arr[],
                                int n, int k)
{
    // Traverse the array from end
    // to start
    for (int i = n - 1; i >= 0; i--)
    {
        // if k is greater than
        // arr[i] then subtract
        // it from k
        if (k >= arr[i])
            k -= arr[i];
    }
 
    // If there is any subsequence
    // whose sum is equal to k
    if (k != 0)
        return false;
    else
        return true;
}
 
// Driver code
public static void main (String[] args)
{
 
    int A[] = { 1, 3, 7, 15, 31 };
    int n = A.length;
    System.out.println(CheckForSequence(A, n, 18) ?
                                            "True": "False");
}
}
 
// This code is contributed by jit_t

Python3

# Python3 implementation of above approach
 
# Function to check whether sum of any set
# of the array element is equal
# to k or not
def CheckForSequence(arr, n, k) :
 
    # Traverse the array from end
    # to start
    for i in range(n - 1, -1, -1) :
        # if k is greater than
        # arr[i] then subtract
        # it from k
        if (k >= arr[i]) :
            k -= arr[i];
 
    # If there is any subsequence
    # whose sum is equal to k
    if (k != 0) :
        return False;
    else :
        return True;
 
# Driver code
if __name__ == "__main__" :
 
    A = [ 1, 3, 7, 15, 31 ];
    n = len(A);
     
    if (CheckForSequence(A, n, 18)) :
        print(True)
    else :
        print(False)
         
# This code is contributed by AnkitRai01

C#

// C# implementation of above approach
using System;
 
class GFG
{
     
// Function to check whether
// sum of any set of the array element
// is equal to k or not
static bool CheckForSequence(int []arr,
                                int n, int k)
{
    // Traverse the array from end
    // to start
    for (int i = n - 1; i >= 0; i--)
    {
        // if k is greater than
        // arr[i] then subtract
        // it from k
        if (k >= arr[i])
            k -= arr[i];
    }
 
    // If there is any subsequence
    // whose sum is equal to k
    if (k != 0)
        return false;
    else
        return true;
}
 
// Driver code
public static void Main ()
{
 
    int []A = { 1, 3, 7, 15, 31 };
    int n = A.Length;
    Console.WriteLine(CheckForSequence(A, n, 18) ?
                                            "True": "False");
}
}
 
// This code is contributed by anuj_67..

Javascript

// Javascript program to implement the above approach
 
// Function to check whether sum of any set
// of the array element is equal
// to k or not
function CheckForSequence( arr,n,k)
{
   // Traverse the array from end
    // to start
    for (var i = n - 1; i >= 0; i--) {
        // if k is greater than
        // arr[i] then subtract
        // it from k
        if (k >= arr[i])
            k -= arr[i];
    }
 
    // If there is any subsequence
    // whose sum is equal to k
    if (k != 0)
        return false;
    else
        return true;
}
 
var A = [ 1, 3, 7, 15, 31 ];
var n = A.length;
document.write( (CheckForSequence(A, n, 18) ? "True": "False") +"<br>");
 
//This code is contributed by SoumikModnal

输出：

True

时间复杂度：O(N)