将全1放在单个索引所需的最小步骤数

给定大小为N的二进制数组A [] ，可以将所有1 s移动到其相邻位置，任务是打印大小为N的数组res [] ，其中res [i]包含执行以下操作所需的最少步骤数：将所有1 s移至第i^个索引。

例子：

Input: A[] = {1, 0, 1, 0}
Output: {2, 2, 2, 4}
Explanation: 
For i = 0, moving all 1s to 0^th index requires 2 steps, i.e abs(0 – 0) + abs(0 – 2) = 2.
For i = 1, moving all 1s to 1^st index requires 2 steps, i.e abs(1 – 0) + abs(1 – 2) = 2.
For i = 2, moving all 1s to 2^nd index requires 2 steps, i.e abs(2 – 0) + abs(2 – 2) = 2.
For i = 3, moving all 1’s to 3^rd index requires 4 steps, i.e abs(3 – 0) + abs(3 – 2) = 4.
Hence, res[] is {2, 2, 2, 4}

Input: A[] = {0, 0, 1, 0, 0, 1}
Output: {7, 5, 3, 3, 3, 3}
Explanation:
For i = 0, moving all 1s to 0^th index requires 7 steps, i.e abs(2 – 0) + abs(5 – 0) = 7.
For i = 1, moving all 1s to 1^st index requires 5 steps, i.e abs(2 – 1) + abs(5 – 1) = 5.
For i = 2, moving all 1s to 2^nd index requires 3 steps, i.e abs(2 – 2) + abs(5 – 2) = 3.
For i = 3, moving all 1s to 3rd index will take 3 steps, i.e abs(2 – 2) + abs(5 – 3) = 3.
For i = 4, moving all 1s to 4^th index will take 3 steps, i.e abs(2 – 4) + abs(5 – 4) = 3.
For i = 5, moving all 1s to 5^th index will take 3 steps, i.e abs(2 – 5) + abs(5 – 5) = 3.
Hence, res[] is {7, 5, 3, 3, 3, 3}
 

天真的方法：请按照以下步骤解决问题：

1. 遍历数组。
2. 对于每个^第i^个索引，计算将所有1 s移到第i^个索引所需的步骤数。
3. 使用变量i遍历[0，N – 1]范围。
   • 初始化步骤= 0。
   • 使用变量j在[0，N – 1]范围内迭代。
   • 如果A [j]等于1，则将abs(i – j)添加到步骤中。
4. 打印最小步数。

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，我们的想法是使用Prefix Sum。请按照以下步骤解决此问题：

1. 遍历数组。
2. 初始化一个数组，例如left []，并初始化count = A [0]。
3. 在[1，N – 1]范围内进行迭代，并更新left [i] = left [i – 1] + count ，其中count是i左侧1 s的数量。
4. 初始化一个数组，例如right []，并初始化count = A [N – 1]。
5. 在[N – 2，0]范围内进行迭代，并更新right [i] = right [i + 1] + count ，其中count是i右侧1 s的数量。
6. 计算并更新res []中的最终答案，其中res [i]是左右两侧步长的总和，即res [i] = right [i] + left [i]
7. 打印数组res [] 。

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
#include 
using namespace std;
 
// Function to print minimum steps
// required to shift all 1s to a
// single index in a binary array
void minsteps(vector& A)
{
    // Size of array
    int n = A.size();
 
    // Used to store cummulative sum
    vector left(n, 0), right(n, 0), res(n, 0);
 
    // Initialize count
    int count = A[0];
 
    // Traverse the array in
    // forward direction
    for (int i = 1; i < n; i++) {
        // Steps needed to store all
        // previous ones to ith index
        left[i] = left[i - 1] + count;
 
        // Count number of 1s
        // present till i-th index
        count += A[i];
    }
 
    // Initialize count
    count = A[n - 1];
 
    // Traverse the array in
    // backward direction
    for (int i = n - 2; i >= 0; i--) {
        // Steps needed to store all 1s to
        // the right of i at current index
        right[i] = right[i + 1] + count;
 
        // Count number of 1s
        // present after i-th index
        count += A[i];
    }
 
    // Print the number of steps required
    for (int i = 0; i < n; i++) {
        res[i] = left[i] + right[i];
        cout << res[i] << " ";
    }
    cout << "\n";
}
 
// Driver Code
int main()
{
    vector A = { 1, 0, 1, 0 };
    minsteps(A);
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG
{
 
  // Function to print minimum steps
  // required to shift all 1s to a
  // single index in a binary array
  static void minsteps(int[] A)
  {
     
    // Size of array
    int n = A.length;
 
    // Used to store cummulative sum
    int[] left = new int [n];
    Arrays.fill(left, 0);
    int[] right = new int [n];
    Arrays.fill(right, 0);
    int[] res = new int [n];
    Arrays.fill(res, 0);
 
    // Initialize count
    int count = A[0];
 
    // Traverse the array in
    // forward direction
    for (int i = 1; i < n; i++) {
      // Steps needed to store all
      // previous ones to ith index
      left[i] = left[i - 1] + count;
 
      // Count number of 1s
      // present till i-th index
      count += A[i];
    }
 
    // Initialize count
    count = A[n - 1];
 
    // Traverse the array in
    // backward direction
    for (int i = n - 2; i >= 0; i--) {
      // Steps needed to store all 1s to
      // the right of i at current index
      right[i] = right[i + 1] + count;
 
      // Count number of 1s
      // present after i-th index
      count += A[i];
    }
 
    // Print the number of steps required
    for (int i = 0; i < n; i++) {
      res[i] = left[i] + right[i];
      System.out.print(res[i] + " ");
    }
    System.out.println();
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int[] A = { 1, 0, 1, 0 };
    minsteps(A);
  }
}
 
// This code is contributed by souravghosh0416.

Python3

# Python3 implementation of
# the above approach
 
# Function to prminimum steps
# required to shift all 1s to a
# single index in a binary array
def minsteps(A):
   
    # Size of array
    n = len(A)
 
    # Used to store cummulative sum
    left, right, res =[0]*n, [0]*n, [0]*n
 
    # Initialize count
    count = A[0]
 
    # Traverse the array in
    # forward direction
    for i in range(1, n):
       
        # Steps needed to store all
        # previous ones to ith index
        left[i] = left[i - 1] + count
 
        # Count number of 1s
        # present till i-th index
        count += A[i]
 
    # Initialize count
    count = A[n - 1]
 
    # Traverse the array in
    # backward direction
    for i in range(n - 2, -1, -1):
       
        # Steps needed to store all 1s to
        # the right of i at current index
        right[i] = right[i + 1] + count
 
        # Count number of 1s
        # present after i-th index
        count += A[i]
 
    # Prthe number of steps required
    for i in range(n):
        res[i] = left[i] + right[i]
        print(res[i], end = " ")
    print()
 
# Driver Code
if __name__ == '__main__':
    A = [1, 0, 1, 0]
    minsteps(A)
 
# This code is contributed by mohit kumar 29.

C#

// C# Program to implement
// the above approach
using System;
class GFG
{
   
  // Function to print minimum steps
  // required to shift all 1s to a
  // single index in a binary array
  static void minsteps(int[] A)
  {
 
    // Size of array
    int n = A.Length;
 
    // Used to store cummulative sum
    int[] left = new int [n];
    for (int i = 1; i < n; i++) {
      left[i] = 0;
    }
 
    int[] right = new int [n];
    for (int i = 1; i < n; i++) {
      right[i] = 0;
    }
 
    int[] res = new int [n];
    for (int i = 1; i < n; i++) {
      res[i] = 0;
    }
 
    // Initialize count
    int count = A[0];
 
    // Traverse the array in
    // forward direction
    for (int i = 1; i < n; i++) {
      // Steps needed to store all
      // previous ones to ith index
      left[i] = left[i - 1] + count;
 
      // Count number of 1s
      // present till i-th index
      count += A[i];
    }
 
    // Initialize count
    count = A[n - 1];
 
    // Traverse the array in
    // backward direction
    for (int i = n - 2; i >= 0; i--) {
      // Steps needed to store all 1s to
      // the right of i at current index
      right[i] = right[i + 1] + count;
 
      // Count number of 1s
      // present after i-th index
      count += A[i];
    }
 
    // Print the number of steps required
    for (int i = 0; i < n; i++) {
      res[i] = left[i] + right[i];
      Console.Write(res[i] + " ");
    }
    Console.WriteLine();
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int[] A = { 1, 0, 1, 0 };
    minsteps(A);
  }
}
 
// This code is contributed by susmitakundugoaldanga.

输出：

2 2 2 4

时间复杂度： O(N)
辅助空间： O(N)