# Maximum distance between two 1s in a Binary Array in a given range

Given a binary array of size N and a range in [l, r], the task is to find the maximum distance between two 1s in this given range.

Examples:

Input: arr = {1, 0, 0, 1}, l = 0, r = 3
Output: 3
In the given range from 0 to 3, first 1 lies at index 0 and last at index 3.
Hence, maximum distance = 3 – 0 = 3.

Input: arr = {1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0}, l = 3, r = 9
Output: 6

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

Approach: We will create a segment tree to solve this.

1. Each node in the segment tree will have the index of leftmost 1 as well as rightmost 1 and an integer containing the maximum distance between any elements with value 1 in a subarray {l, r}.
2. Now, in this segment tree we can merge left and right nodes as below:
• If left node is not valid, return right node.
• If right node is not valid, return left node.
• A node is valid iff it contains at least one 0, or at least one 1.
• If both left and right nodes are valid then:
Let,

• l1 = leftmost index of 1 (-1 if 0 doesn’t exist in that interval)
• r1 = rightmost index of 1 (-1 if 0 doesn’t exist in that interval)
• max1 = maximum distance between two 1’s

then,

• Value of max1 in merged node will be maximum of maximum of value of max1 in left and right node, and difference between rightmost index of 1 in right node and leftmost index of 1 in left node.
• Value of l1 in merged node will be l1 of left node if it is not -1, else l1 of right node.
• Value of r1 in merged node will be r1 of right node if it is not -1, else r1 of left node.
3. Then, finally to find the answer we just need to call query function for the given range {l, r}.

Below is the implementation of the above approach:

## C++

 // C++ program to find the maximum // distance between two elements // with value 1 within a subarray (l, r)    #include using namespace std;    // Structure for each node // in the segment tree struct node {     int l1, r1;     int max1; } seg[100001];    // A utility function for // merging two nodes node task(node l, node r) {     node x;        x.l1 = (l.l1 != -1) ? l.l1 : r.l1;     x.r1 = (r.r1 != -1) ? r.r1 : l.r1;     x.max1 = max(l.max1, r.max1);        if (l.l1 != -1 && r.r1 != -1)         x.max1 = max(x.max1, r.r1 - l.l1);        return x; }    // A recursive function that constructs // Segment Tree for given string void build(int qs, int qe, int ind, int arr[]) {     // If start is equal to end then     // insert the array element     if (qs == qe) {         if (arr[qs] == 1) {             seg[ind].l1 = seg[ind].r1 = qs;             seg[ind].max1 = INT_MIN;         }         else {             seg[ind].l1 = seg[ind].r1 = -1;             seg[ind].max1 = INT_MIN;         }            return;     }     int mid = (qs + qe) >> 1;        // Build the segment tree     // for range qs to mid     build(qs, mid, ind << 1, arr);        // Build the segment tree     // for range mid+1 to qe     build(mid + 1, qe, ind << 1 | 1, arr);        // merge the two child nodes     // to obtain the parent node     seg[ind] = task(         seg[ind << 1],         seg[ind << 1 | 1]); }    // Query in a range qs to qe node query(int qs, int qe,            int ns, int ne, int ind) {     node x;     x.l1 = x.r1 = -1;     x.max1 = INT_MIN;        // If the range lies in this segment     if (qs <= ns && qe >= ne)         return seg[ind];        // If the range is out of the bounds     // of this segment     if (ne < qs || ns > qe || ns > ne)         return x;        // Else query for the right and left     // child node of this subtree     // and merge them     int mid = (ns + ne) >> 1;        node l = query(qs, qe, ns,                    mid, ind << 1);     node r = query(qs, qe,                    mid + 1, ne,                    ind << 1 | 1);        x = task(l, r);     return x; }    // Driver code int main() {        int arr[] = { 1, 1, 0,                   1, 0, 1,                   0, 1, 0,                   1, 0, 1,                   1, 0 };     int n = sizeof(arr) / sizeof(arr[0]);     int l = 3, r = 9;        // Build the segment tree     build(0, n - 1, 1, arr);        // Query in range 3 to 9     node ans = query(l, r, 0, n - 1, 1);     cout << ans.max1 << "\n";        return 0; }

## Java

 // Java program to find the maximum // distance between two elements // with value 1 within a subarray (l, r) import java.util.*;    class GFG{     // Structure for each node // in the segment tree static class node {     int l1, r1;     int max1; } static node []seg = new node[100001];     // A utility function for // merging two nodes static node task(node l, node r) {     node x = new node();         x.l1 = (l.l1 != -1) ? l.l1 : r.l1;     x.r1 = (r.r1 != -1) ? r.r1 : l.r1;     x.max1 = Math.max(l.max1, r.max1);         if (l.l1 != -1 && r.r1 != -1)         x.max1 = Math.max(x.max1, r.r1 - l.l1);         return x; }     // A recursive function that constructs // Segment Tree for given String static void build(int qs, int qe, int ind, int arr[]) {     // If start is equal to end then     // insert the array element     if (qs == qe) {         if (arr[qs] == 1) {             seg[ind].l1 = seg[ind].r1 = qs;             seg[ind].max1 = Integer.MIN_VALUE;         }         else {             seg[ind].l1 = seg[ind].r1 = -1;             seg[ind].max1 = Integer.MIN_VALUE;         }             return;     }     int mid = (qs + qe) >> 1;         // Build the segment tree     // for range qs to mid     build(qs, mid, ind << 1, arr);         // Build the segment tree     // for range mid+1 to qe     build(mid + 1, qe, ind << 1 | 1, arr);         // merge the two child nodes     // to obtain the parent node     seg[ind] = task(         seg[ind << 1],         seg[ind << 1 | 1]); }     // Query in a range qs to qe static node query(int qs, int qe,            int ns, int ne, int ind) {     node x = new node();     x.l1 = x.r1 = -1;     x.max1 = Integer.MIN_VALUE;         // If the range lies in this segment     if (qs <= ns && qe >= ne)         return seg[ind];         // If the range is out of the bounds     // of this segment     if (ne < qs || ns > qe || ns > ne)         return x;         // Else query for the right and left     // child node of this subtree     // and merge them     int mid = (ns + ne) >> 1;         node l = query(qs, qe, ns,                    mid, ind << 1);     node r = query(qs, qe,                    mid + 1, ne,                    ind << 1 | 1);         x = task(l, r);     return x; }     // Driver code public static void main(String[] args) {                for(int i= 0; i < 100001; i++)         seg[i] = new node();     int arr[] = { 1, 1, 0,                   1, 0, 1,                   0, 1, 0,                   1, 0, 1,                   1, 0 };     int n = arr.length;     int l = 3, r = 9;         // Build the segment tree     build(0, n - 1, 1, arr);         // Query in range 3 to 9     node ans = query(l, r, 0, n - 1, 1);     System.out.print(ans.max1+ "\n");     } }    // This code is contributed by Rajput-Ji

## C#

 // C# program to find the maximum // distance between two elements // with value 1 within a subarray (l, r) using System;    class GFG{      // Structure for each node // in the segment tree class node {     public int l1, r1;     public int max1; } static node []seg = new node[100001];      // A utility function for // merging two nodes static node task(node l, node r) {     node x = new node();          x.l1 = (l.l1 != -1) ? l.l1 : r.l1;     x.r1 = (r.r1 != -1) ? r.r1 : l.r1;     x.max1 = Math.Max(l.max1, r.max1);          if (l.l1 != -1 && r.r1 != -1)         x.max1 = Math.Max(x.max1, r.r1 - l.l1);          return x; }      // A recursive function that constructs // Segment Tree for given String static void build(int qs, int qe, int ind, int []arr) {     // If start is equal to end then     // insert the array element     if (qs == qe) {         if (arr[qs] == 1) {             seg[ind].l1 = seg[ind].r1 = qs;             seg[ind].max1 = int.MinValue;         }         else {             seg[ind].l1 = seg[ind].r1 = -1;             seg[ind].max1 = int.MinValue;         }              return;     }     int mid = (qs + qe) >> 1;          // Build the segment tree     // for range qs to mid     build(qs, mid, ind << 1, arr);          // Build the segment tree     // for range mid+1 to qe     build(mid + 1, qe, ind << 1 | 1, arr);          // merge the two child nodes     // to obtain the parent node     seg[ind] = task(         seg[ind << 1],         seg[ind << 1 | 1]); }      // Query in a range qs to qe static node query(int qs, int qe,            int ns, int ne, int ind) {     node x = new node();     x.l1 = x.r1 = -1;     x.max1 = int.MinValue;          // If the range lies in this segment     if (qs <= ns && qe >= ne)         return seg[ind];          // If the range is out of the bounds     // of this segment     if (ne < qs || ns > qe || ns > ne)         return x;          // Else query for the right and left     // child node of this subtree     // and merge them     int mid = (ns + ne) >> 1;          node l = query(qs, qe, ns,                    mid, ind << 1);     node r = query(qs, qe,                    mid + 1, ne,                    ind << 1 | 1);          x = task(l, r);     return x; }      // Driver code public static void Main(String[] args) {                  for(int i = 0; i < 100001; i++)         seg[i] = new node();     int []arr = { 1, 1, 0,                   1, 0, 1,                   0, 1, 0,                   1, 0, 1,                   1, 0 };     int n = arr.Length;     int l = 3, r = 9;          // Build the segment tree     build(0, n - 1, 1, arr);          // Query in range 3 to 9     node ans = query(l, r, 0, n - 1, 1);     Console.Write(ans.max1+ "\n"); } }     // This code is contributed by Princi Singh

Output:

6

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