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

• Last Updated : 22 Jun, 2021

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:
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:

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
• Value of max1 in merged node will be 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`

## Python3

 `# Python3 program to find the maximum``# distance between two elements``# with value 1 within a subarray (l, r)``import` `sys` `# Structure for each node``# in the segment tree``class` `node():``    ` `    ``def` `__init__(``self``):``        ` `        ``self``.l1 ``=` `0``        ``self``.r1 ``=` `0``        ``self``.max1 ``=` `0``        ` `seg ``=` `[node() ``for` `i ``in` `range``(``100001``)]` `# A utility function for``# merging two nodes``def` `task(l, r):``    ` `    ``x ``=` `node()`` ` `    ``x.l1 ``=` `l.l1 ``if` `(l.l1 !``=` `-``1``) ``else` `r.l1``    ``x.r1 ``=` `r.r1 ``if` `(r.r1 !``=` `-``1``)  ``else` `l.r1`` ` `    ``x.max1 ``=` `max``(l.max1, r.max1)`` ` `    ``# If both the nodes are valid``    ``if` `(l.r1 !``=` `-``1` `and` `r.l1 !``=` `-``1``):``        ``x.max1 ``=` `max``(x.max1, r.r1 ``-` `l.l1)`` ` `    ``return` `x` `# A recursive function that constructs``# Segment Tree for given string``def` `build(qs, qe, ind, 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 ``=` `-``sys.maxsize``            ` `        ``else``:``            ``seg[ind].l1 ``=` `seg[ind].r1 ``=` `-``1``            ``seg[ind].max1 ``=` `-``sys.maxsize`` ` `        ``return`` ` `    ``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``def` `query(qs, qe, ns, ne, ind):` `    ``x ``=` `node()``    ``x.l1 ``=` `x.r1 ``=` `-``1``    ``x.max1 ``=` `-``sys.maxsize`` ` `    ``# If the range lies in this segment``    ``if` `(qs <``=` `ns ``and` `qe >``=` `ne):``        ``return` `seg[ind]`` ` `    ``# If the range is out of the bounds``    ``# of this segment``    ``if` `(ne < qs ``or` `ns > qe ``or` `ns > ne):``        ``return` `x`` ` `    ``# Else query for the right and left``    ``# child node of this subtree``    ``# and merge them``    ``mid ``=` `(ns ``+` `ne) >> ``1`` ` `    ``l ``=` `query(qs, qe, ns, mid, ind << ``1``)``    ``r ``=` `query(qs, qe, mid ``+` `1``, ne, ind << ``1` `| ``1``)`` ` `    ``x ``=` `task(l, r)``    ` `    ``return` `x``    ` `# Driver code``if` `__name__``=``=``"__main__"``:``    ` `    ``arr ``=` `[ ``1``, ``1``, ``0``, ``1``, ``0``, ``1``, ``0``,``            ``1``, ``0``, ``1``, ``0``, ``1``, ``1``, ``0` `]`` ` `    ``n ``=` `len``(arr)``    ` `    ``l ``=` `3``    ``r ``=` `9``    ` `    ``# Build the segment tree``    ``build(``0``, n ``-` `1``, ``1``, arr)``    ` `    ``# Query in range 3 to 9``    ``ans ``=` `query(l, r, ``0``, n ``-` `1``, ``1``)``    ` `    ``print``(ans.max1)` `# This code is contributed by rutvik_56`

## 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`

## Javascript

 ``
Output:
`6`

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