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📜  不能被X整除的最长子数组

📅  最后修改于: 2021-05-14 02:30:23             🧑  作者: Mango

给定一个数组arr []和一个整数X ,任务是打印最长的子数组,以使它的元素之和不被X整除。如果不存在这样的子数组,则打印“ -1”
注意:如果存在多个具有给定属性的子数组,请打印其中的任何一个。
例子:

天真的方法:解决问题的最简单方法是生成所有可能的子数组并继续计算其和。如果发现任何子数组的总和不能被X整除,则将长度与获得的最大长度( maxm )进行比较,并相应地更新maxm并更新子数组开始索引结束索引。最后,打印具有已存储的开始索引和结束索引的子数组。如果没有这样的子数组,则打印“ -1”
时间复杂度: O(N 2 )
辅助空间: O(1)

高效方法:为了优化上述方法,我们将找到前缀和后缀数组总和。请按照以下步骤操作:

  • 生成前缀和数组后缀和数组
  • 使用两个指针从[0,N – 1]进行迭代并选择每个不能被X整除的索引处元素的前缀和后缀和。存储子数组的开始索引和结束索引
  • 完成上述步骤后,如果存在一个总和不能被X整除的子数组,则打印具有已存储的开始索引和结束索引的子数组。
  • 如果没有这样的子数组,则打印“ -1”

下面是上述方法的实现:

C++
#include 
// C++ Program to implement
// the above approach
#include 
using namespace std;
  
// Function to print the longest
// subarray with sum of elements
// not divisible by X
void max_length(int N, int x,
                vector& v)
{
    int i, a;
  
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    vector preff, suff;
    int ct = 0;
  
    for (i = 0; i < N; i++) {
  
        a = v[i];
  
        // If array element is
        // divisibile by x
        if (a % x == 0) {
  
            // Increase count
            ct += 1;
        }
    }
  
    // If all the array elements
    // are divisible by x
    if (ct == N) {
  
        // No subarray possible
        cout << -1 << endl;
        return;
    }
  
    // Reverse v to calculate the
    // suffix sum
    reverse(v.begin(), v.end());
  
    suff.push_back(v[0]);
  
    // Calculate the suffix sum
    for (i = 1; i < N; i++) {
        suff.push_back(v[i]
                       + suff[i - 1]);
    }
  
    // Reverse to original form
    reverse(v.begin(), v.end());
  
    // Reverse the suffix sum array
    reverse(suff.begin(), suff.end());
  
    preff.push_back(v[0]);
  
    // Calculate the prefix sum
    for (i = 1; i < N; i++) {
        preff.push_back(v[i]
                        + preff[i - 1]);
    }
  
    int ans = 0;
  
    // Stores the starting index
    // of required subarray
    int lp = 0;
  
    // Stores the ending index
    // of required subarray
    int rp = N - 1;
  
    for (i = 0; i < N; i++) {
  
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff[i] % x != 0
            && (ans < (N - 1))) {
  
            lp = i;
            rp = N - 1;
  
            // Update the answer
            ans = max(ans, N - i);
        }
  
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff[i] % x != 0
            && (ans < (i + 1))) {
  
            lp = 0;
            rp = i;
  
            // Update the answer
            ans = max(ans, i + 1);
        }
    }
  
    // Print the longest subarray
    for (i = lp; i <= rp; i++) {
        cout << v[i] << " ";
    }
}
  
// Driver Code
int main()
{
    int x = 3;
  
    vector v = { 1, 3, 2, 6 };
    int N = v.size();
  
    max_length(N, x, v);
  
    return 0;
}


Java
// Java program to implement
// the above approach
import java.util.*;
  
class GFG{
  
// Function to print the longest
// subarray with sum of elements
// not divisible by X
static void max_length(int N, int x,
                       int []v)
{
    int i, a;
  
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    List preff = new Vector();
    List suff = new Vector();
      
    int ct = 0;
  
    for(i = 0; i < N; i++) 
    {
        a = v[i];
  
        // If array element is
        // divisibile by x
        if (a % x == 0)
        {
              
            // Increase count
            ct += 1;
        }
    }
  
    // If all the array elements
    // are divisible by x
    if (ct == N) 
    {
          
        // No subarray possible
        System.out.print(-1 + "\n");
        return;
    }
  
    // Reverse v to calculate the
    // suffix sum
    v = reverse(v);
  
    suff.add(v[0]);
  
    // Calculate the suffix sum
    for(i = 1; i < N; i++)
    {
        suff.add(v[i] + suff.get(i - 1));
    }
  
    // Reverse to original form
    v = reverse(v);
  
    // Reverse the suffix sum array
    Collections.reverse(suff);
  
    preff.add(v[0]);
  
    // Calculate the prefix sum
    for(i = 1; i < N; i++)
    {
        preff.add(v[i] + preff.get(i - 1));
    }
  
    int ans = 0;
  
    // Stores the starting index
    // of required subarray
    int lp = 0;
  
    // Stores the ending index
    // of required subarray
    int rp = N - 1;
  
    for(i = 0; i < N; i++)
    {
          
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff.get(i) % x != 0 &&
           (ans < (N - 1))) 
        {
            lp = i;
            rp = N - 1;
  
            // Update the answer
            ans = Math.max(ans, N - i);
        }
  
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff.get(i) % x != 0 &&
           (ans < (i + 1)))
        {
            lp = 0;
            rp = i;
  
            // Update the answer
            ans = Math.max(ans, i + 1);
        }
    }
  
    // Print the longest subarray
    for(i = lp; i <= rp; i++)
    {
        System.out.print(v[i] + " ");
    }
}
  
static int[] reverse(int a[]) 
{
    int i, n = a.length, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
  
// Driver Code
public static void main(String[] args)
{
    int x = 3;
    int []v = { 1, 3, 2, 6 };
    int N = v.length;
  
    max_length(N, x, v);
}
}
  
// This code is contributed by PrinciRaj1992


Python3
# Python3 program to implement 
# the above approach 
  
# Function to print the longest 
# subarray with sum of elements 
# not divisible by X 
def max_length(N, x, v):
      
    # Pref[] stores the prefix sum 
    # Suff[] stores the suffix sum 
    preff, suff = [], []
    ct = 0
      
    for i in range(N):
        a = v[i]
          
        # If array element is 
        # divisibile by x 
        if a % x == 0:
              
            # Increase count 
            ct += 1
              
    # If all the array elements 
    # are divisible by x 
    if ct == N:
          
        # No subarray possible 
        print(-1)
        return
      
    # Reverse v to calculate the 
    # suffix sum 
    v.reverse()
      
    suff.append(v[0])
      
    # Calculate the suffix sum 
    for i in range(1, N):
        suff.append(v[i] + suff[i - 1])
          
    # Reverse to original form 
    v.reverse()
      
    # Reverse the suffix sum array
    suff.reverse()
      
    preff.append(v[0])
      
    # Calculate the prefix sum
    for i in range(1, N):
        preff.append(v[i] + preff[i - 1])
          
    ans = 0
      
    # Stores the starting index 
    # of required subarray 
    lp = 0
      
    # Stores the ending index 
    # of required subarray 
    rp = N - 1
      
    for i in range(N):
          
        # If suffix sum till i-th 
        # index is not divisible by x 
        if suff[i] % x != 0 and ans < N - 1:
            lp = i
            rp = N - 1
              
            # Update the answer
            ans = max(ans, N - i)
              
        # If prefix sum till i-th 
        # index is not divisible by x 
        if preff[i] % x != 0 and ans < i + 1:
            lp = 0
            rp = i
              
            # Update the answer
            ans = max(ans, i + 1)
              
    # Print the longest subarray
    for i in range(lp, rp + 1):
        print(v[i], end = " ")
          
# Driver code
x = 3
v = [ 1, 3, 2, 6 ]
N = len(v)
  
max_length(N, x, v)
  
# This code is contributed by Stuti Pathak


C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to print the longest
// subarray with sum of elements
// not divisible by X
static void max_length(int N, int x,
                       int []v)
{
    int i, a;
  
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    List preff = new List();
    List suff = new List();
      
    int ct = 0;
  
    for(i = 0; i < N; i++) 
    {
        a = v[i];
  
        // If array element is
        // divisibile by x
        if (a % x == 0)
        {
              
            // Increase count
            ct += 1;
        }
    }
  
    // If all the array elements
    // are divisible by x
    if (ct == N) 
    {
          
        // No subarray possible
        Console.Write(-1 + "\n");
        return;
    }
  
    // Reverse v to calculate the
    // suffix sum
    v = reverse(v);
  
    suff.Add(v[0]);
  
    // Calculate the suffix sum
    for(i = 1; i < N; i++)
    {
        suff.Add(v[i] + suff[i - 1]);
    }
  
    // Reverse to original form
    v = reverse(v);
  
    // Reverse the suffix sum array
    suff.Reverse();
  
    preff.Add(v[0]);
  
    // Calculate the prefix sum
    for(i = 1; i < N; i++)
    {
        preff.Add(v[i] + preff[i - 1]);
    }
  
    int ans = 0;
  
    // Stores the starting index
    // of required subarray
    int lp = 0;
  
    // Stores the ending index
    // of required subarray
    int rp = N - 1;
  
    for(i = 0; i < N; i++)
    {
          
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff[i] % x != 0 &&
               (ans < (N - 1))) 
        {
            lp = i;
            rp = N - 1;
  
            // Update the answer
            ans = Math.Max(ans, N - i);
        }
  
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff[i] % x != 0 &&
                (ans < (i + 1)))
        {
            lp = 0;
            rp = i;
  
            // Update the answer
            ans = Math.Max(ans, i + 1);
        }
    }
  
    // Print the longest subarray
    for(i = lp; i <= rp; i++)
    {
        Console.Write(v[i] + " ");
    }
}
  
static int[] reverse(int []a) 
{
    int i, n = a.Length, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
  
// Driver Code
public static void Main(String[] args)
{
    int x = 3;
    int []v = { 1, 3, 2, 6 };
    int N = v.Length;
  
    max_length(N, x, v);
}
}
  
// This code is contributed by PrinciRaj1992


输出:
3 2 6

时间复杂度: O(N)
辅助空间: O(N)