# Distance of closest zero to every element

Given an array of n integers, for each element, print the distance to the closest zero. Array has a minimum of 1 zero in it.

Examples:

Input: 5 6 0 1 -2 3 4
Output: 2 1 0 1 2 3 4
Explanation : The nearest 0(indexed 2) to
5(indexed 0) is at a distance of 2, so we
print 2. Same is done for the rest of elements.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: A naive approach is, for every element, slide towards left and find out the nearest 0 and again slide towards the right to find out the nearest zero if any, and print the minimum of both the distances. It will be space efficient but the time complexity will be high as we have to iterate for every element till we find the 0, and in worst case we may not find in one direction.
Time Complexity: O(n^2)
Auxiliary Space: O(1)

Efficient Approach: An efficient approach is to use sliding window technique two time. One is traversing from right to left and other from left right.
Initialize ans[0] with a max value. Iterate over array from left to right. If value in current position is 0, then set distance to 0, otherwise increase distance by 1. In each step, write value of distance to the answer array.
Do the same thing but going from right to left. This will find closest zero to the right. Now we should store the minimum of current value of distance and value that’s already in answer array.
Below is the implementation of the above approach.

## C++

 // CPP program to find closest 0 for every element #include using namespace std;    // Print the distance with zeroes of every element void print_distance(int arr[], int n) {     // initializes an array of size n with 0     int ans[n];     memset(arr, 0, sizeof(arr));        // if first element is 0 then the distance     // will be 0     if (arr[0] == 0)         ans[0] = 0;     else         ans[0] = INT_MAX; // if not 0 then initialize                            // with a maximum value        // traverse in loop from 1 to n and store      // the distance from left     for (int i = 1; i < n; ++i) {            // add 1 to the distance from previous one         ans[i] = ans[i - 1] + 1;            // if the present element is 0 then distance         // will be 0         if (arr[i] == 0)              ans[i] = 0;             }        // if last element is zero then it will be 0 else      // let the answer be what was found when traveled     // form left to right     if (arr[n - 1] == 0)         ans[n - 1] = 0;        // traverse from right to left and store the minimum     // of distance if found from right to left or left     // to right     for (int i = n - 2; i >= 0; --i) {            // store the minimum of distance from left to         // right or right to left         ans[i] = min(ans[i], ans[i + 1] + 1);            // if it is 0 then minimum will always be 0         if (arr[i] == 0)              ans[i] = 0;     }        // print the answer array     for (int i = 0; i < n; ++i)          cout << ans[i] << " ";     }    // driver program to test the above function int main() {     int a[] = { 2, 1, 0, 3, 0, 0, 3, 2, 4 };     int n = sizeof(a) / sizeof(a[0]);     printDistances(a, n);     return 0; }

## Java

 // Java program to find closest // 0 for every element import java.util.Arrays;    class GFG {     // Print the distance with zeroes of every element     static void print_distance(int arr[], int n)     {         // initializes an array of size n with 0         int ans[]=new int[n];         Arrays.fill(ans,0);                // if first element is 0 then the distance         // will be 0         if (arr[0] == 0)             ans[0] = 0;                    // if not 0 then initialize          // with a maximum value             else             ans[0] = +2147483647;                // traverse in loop from 1 to n and store          // the distance from left         for (int i = 1; i < n; ++i)          {                    // add 1 to the distance             // from previous one             ans[i] = ans[i - 1] + 1;                    // if the present element is              // 0 then distance will be 0             if (arr[i] == 0)                  ans[i] = 0;              }                // if last element is zero          // then it will be 0 else          // let the answer be what was         // found when traveled         // form left to right         if (arr[n - 1] == 0)             ans[n - 1] = 0;                // traverse from right to          // left and store the minimum         // of distance if found from          // right to left or left         // to right         for (int i = n - 2; i >= 0; --i)          {                    // store the minimum of distance              // from left to right or right to left             ans[i] = Math.min(ans[i], ans[i + 1] + 1);                    // if it is 0 then minimum              // will always be 0             if (arr[i] == 0)                  ans[i] = 0;         }                // print the answer array         for (int i = 0; i < n; ++i)              System.out.print(ans[i] + " ");      }            // Driver code      public static void main (String[] args)     {         int a[] = { 2, 1, 0, 3, 0, 0, 3, 2, 4 };         int n = a.length;         print_distance(a, n);     } }    // This code is contributed by Anant Agarwal.

## Python3

 # Python3 program to find closest 0 # for every element    # Print the distance with zeroes of  # every element def print_distance(arr, n):            # initializes an array of size n with 0     ans = [0 for i in range(n)]            # if first element is 0 then the      # distance will be 0     if (arr[0] == 0):         ans[0] = 0     else:         ans[0] = 10**9  # if not 0 then initialize                          # with a maximum value        # traverse in loop from 1 to n and      # store the distance from left     for i in range(1, n):            # add 1 to the distance from          # previous one         ans[i] = ans[i - 1] + 1            # if the present element is 0 then          # distance will be 0         if (arr[i] == 0):             ans[i] = 0        # if last element is zero then it will be 0      # else let the answer be what was found when      # traveled form left to right     if (arr[n - 1] == 0):         ans[n - 1] = 0        # traverse from right to left and store      # the minimum of distance if found from      # right to left or left to right     for i in range(n - 2, -1, -1):            # store the minimum of distance from          # left to right or right to left         ans[i] = min(ans[i], ans[i + 1] + 1)            # if it is 0 then minimum will          # always be 0         if (arr[i] == 0):             ans[i] = 0            # print the answer array     for i in ans:         print(i, end = " ")    # Driver Code a = [2, 1, 0, 3, 0, 0, 3, 2, 4] n = len(a) print_distance(a, n)    # This code is contributed  # by Mohit Kumar

## C#

 // C# program to find closest // 0 for every element using System; class GFG {     // Print the distance with zeroes of every element     static void print_distance(int []arr, int n)     {         // initializes an array of size n with 0         int []ans=new int[n];         for(int i = 0; i < n; i++)             ans[i] = 0;            // if first element is 0 then the distance         // will be 0         if (arr[0] == 0)             ans[0] = 0;                    // if not 0 then initialize          // with a maximum value          else             ans[0] = +2147483646;                // traverse in loop from 1 to n and store          // the distance from left         for (int i = 1; i < n; ++i)          {                    // add 1 to the distance             // from previous one             ans[i] = ans[i - 1] + 1;                    // if the present element is              // 0 then distance will be 0             if (arr[i] == 0)                  ans[i] = 0;              }                // if last element is zero          // then it will be 0 else          // let the answer be what was         // found when traveled         // form left to right         if (arr[n - 1] == 0)             ans[n - 1] = 0;                // traverse from right to          // left and store the minimum         // of distance if found from          // right to left or left         // to right         for (int i = n - 2; i >= 0; --i)          {                    // store the minimum of distance              // from left to right or right to left             ans[i] = Math.Min(ans[i], ans[i + 1] + 1);                    // if it is 0 then minimum              // will always be 0             if (arr[i] == 0)                  ans[i] = 0;         }                // print the answer array         for (int i = 0; i < n; ++i)              Console.Write(ans[i] + " ");      }            // Driver code      public static void Main (String[] args)     {         int []a = { 2, 1, 0, 3, 0, 0, 3, 2, 4 };         int n = a.Length;         print_distance(a, n);     } }    // This code is contributed by PrinciRaj1992

## PHP

 = 0; --\$i)     {             // store the minimum of distance from          // left to right or right to left          \$ans[\$i] = min(\$ans[\$i], \$ans[\$i + 1] + 1);             // if it is 0 then minimum will          // always be 0          if (\$arr[\$i] == 0)              \$ans[\$i] = 0;      }         // print the answer array      for (\$i = 0; \$i < \$n; ++\$i)          echo \$ans[\$i] , " ";  }     // Driver Code \$a = array( 2, 1, 0, 3, 0, 0, 3, 2, 4 );  \$n = sizeof(\$a);  print_distance(\$a, \$n);     // This code is contributed by Sachin ?>

Output:

2 1 0 1 0 0 1 2 3

Time complexity: O(n)
Auxiliary Space: O(n)

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