Maximize length of longest subarray consisting of same elements by at most K decrements

Given an array arr[] of size N and an integer K, the task is to find the length of the longest subarray consisting of same elements that can be obtained by decrementing the array elements by 1 at most K times.

Example:

Input: arr[] = { 1, 2, 3 }, K = 1
Output: 2
Explanation:
Decrementing arr[0] by 1 modifies arr[] to { 1, 1, 3 }
The longest subarray with equal elements is { 1, 1 }.
Therefore, the required output is 2.

Input: arr[] = { 1, 7, 3, 4, 5, 6 }, K = 6
Output: 4

Approach: The problem can be solved using Segment tree and Binary Search technique. The idea is to use the following observations:

Total number of decrements operations required to make all array elements of the subarray { arr[start], …, arr[end] } equal
= (?(start, end)) – (end – start + 1) * (min_value)

where, start = index of the starting point of the subarray
end = index of end point of subarray
min_value = smallest value from index i to j
?(start, end) = sum of all elements from index i to j

Follow the steps below to solve the above problem:

1. Initialize a segment tree to calculate the smallest element in a subarray of the array and a prefix sum array to calculate the sum elements of a subarray.
2. Traverse the array, arr[]. For every ith element perform the following operations:
• Initialize two variables say, start = i, end = N – 1 and apply binary search over the range [start, end] to check if the all the elements of the subarray { arr[start], …, arr[end] } can be made equal or not by decrementing at most K operations from the above observations.
• If all the elements of the subarray { arr[start], …, arr[end] } can be made equal by decrementing at most K operations then update start = (start + end) / 2 + 1.
• Otherwise, update end = (start + end) / 2 – 1

3. Finally, print the length of the longest subarray obtained from the above operations.

Below is the implementation of the above approach:

C++

 `// C++ program to implement ` `// the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to construct Segment Tree ` `// to return the minimum element in a range ` `int` `build(``int` `tree[], ``int``* A, ``int` `start, ` `          ``int` `end, ``int` `node) ` `{ ` `    ``// If leaf nodes of ` `    ``// the tree are found ` `    ``if` `(start == end) { ` ` `  `        ``// Update the value in segment ` `        ``// tree from given array ` `        ``tree[node] = A[start]; ` ` `  `        ``return` `tree[node]; ` `    ``} ` ` `  `    ``// Divide left and right subtree ` `    ``int` `mid = (start + end) / 2; ` ` `  `    ``// Stores smallest element in ` `    ``// subarray { arr[start], arr[mid] } ` `    ``int` `X = build(tree, A, start, mid, ` `                  ``2 * node + 1); ` ` `  `    ``// Stores smallest element in ` `    ``// subarray { arr[mid + 1], arr[end] } ` `    ``int` `Y = build(tree, A, mid + 1, ` `                  ``end, 2 * node + 2); ` ` `  `    ``// Stores smallest element in ` `    ``// subarray { arr[start], arr[end] } ` `    ``return` `tree[node] = min(X, Y); ` `} ` ` `  `// Function to find the smallest ` `// element present in a subarray ` `int` `query(``int` `tree[], ``int` `start, ``int` `end, ` `          ``int` `l, ``int` `r, ``int` `node) ` `{ ` `    ``// If elements of the subarray ` `    ``// are not in the range [l, r] ` `    ``if` `(start > r || end < l) ` `        ``return` `INT_MAX; ` ` `  `    ``// If all the elements of the ` `    ``// subarray are in the range [l, r] ` `    ``if` `(start >= l && end <= r) ` `        ``return` `tree[node]; ` ` `  `    ``// Divide tree into left ` `    ``// and right subtree ` `    ``int` `mid = (start + end) / 2; ` ` `  `    ``// Stores smallest element ` `    ``// in left subtree ` `    ``int` `X = query(tree, start, mid, l, ` `                  ``r, 2 * node + 1); ` ` `  `    ``// Stores smallest element in ` `    ``// right subtree ` `    ``int` `Y = query(tree, mid + 1, end, l, ` `                  ``r, 2 * node + 2); ` ` `  `    ``return` `min(X, Y); ` `} ` ` `  `// Function that find length of longest ` `// subarray with all equal elements in ` `// atmost K decrements ` `int` `longestSubArray(``int``* A, ``int` `N, ``int` `K) ` `{ ` `    ``// Stores length of longest subarray ` `    ``// with all equal elements in atmost ` `    ``// K decrements. ` `    ``int` `res = 1; ` ` `  `    ``// Store the prefix sum array ` `    ``int` `preSum[N + 1]; ` ` `  `    ``// Calculate the prefix sum array ` `    ``preSum[0] = A[0]; ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``preSum[i + 1] = preSum[i] + A[i]; ` ` `  `    ``int` `tree[4 * N + 5]; ` ` `  `    ``// Build the segment tree ` `    ``// for range min query ` `    ``build(tree, A, 0, N - 1, 0); ` ` `  `    ``// Traverse the array ` `    ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `        ``// Stores start index ` `        ``// of the subarray ` `        ``int` `start = i; ` ` `  `        ``// Stores end index ` `        ``// of the subarray ` `        ``int` `end = N - 1; ` ` `  `        ``int` `mid; ` ` `  `        ``// Stores end index of ` `        ``// the longest subarray ` `        ``int` `max_index = i; ` ` `  `        ``// Performing the binary search ` `        ``// to find the endpoint ` `        ``// for the selected range ` `        ``while` `(start <= end) { ` ` `  `            ``// Find the mid for binary search ` `            ``mid = (start + end) / 2; ` ` `  `            ``// Find the smallest element in ` `            ``// range [i, mid] using Segment Tree ` `            ``int` `min_element ` `                ``= query(tree, 0, N - 1, i, mid, 0); ` ` `  `            ``// Stores total sum of subarray ` `            ``// after K decrements ` `            ``int` `expected_sum ` `                ``= (mid - i + 1) * min_element; ` ` `  `            ``// Stores sum of elements of ` `            ``// subarray before K decrements ` `            ``int` `actual_sum ` `                ``= preSum[mid + 1] - preSum[i]; ` ` `  `            ``// If subarray found with ` `            ``// all equal elements ` `            ``if` `(actual_sum - expected_sum <= K) { ` ` `  `                ``// Update start ` `                ``start = mid + 1; ` ` `  `                ``// Update max_index ` `                ``max_index = max(max_index, mid); ` `            ``} ` ` `  `            ``// If false, it means that ` `            ``// the selected range is invalid ` `            ``else` `{ ` ` `  `                ``// Update end ` `                ``end = mid - 1; ` `            ``} ` `        ``} ` ` `  `        ``// Store the length of longest subarray ` `        ``res = max(res, max_index - i + 1); ` `    ``} ` ` `  `    ``// Return result ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 7, 3, 4, 5, 6 }; ` `    ``int` `k = 6; ` `    ``int` `n = 6; ` `    ``cout << longestSubArray(arr, n, k); ` ` `  `    ``return` `0; ` `} `

Java

 `// Java program to implement  ` `// the above approach  ` `import` `java.util.*; ` `  `  `class` `GFG{ ` ` `  `// Function to construct Segment Tree ` `// to return the minimum element in a range ` `static` `int` `build(``int` `tree[], ``int``[] A, ``int` `start, ` `                 ``int` `end, ``int` `node) ` `{ ` `     `  `    ``// If leaf nodes of ` `    ``// the tree are found ` `    ``if` `(start == end)  ` `    ``{ ` `         `  `        ``// Update the value in segment ` `        ``// tree from given array ` `        ``tree[node] = A[start]; ` `  `  `        ``return` `tree[node]; ` `    ``} ` `  `  `    ``// Divide left and right subtree ` `    ``int` `mid = (start + end) / ``2``; ` `  `  `    ``// Stores smallest element in ` `    ``// subarray { arr[start], arr[mid] } ` `    ``int` `X = build(tree, A, start, mid, ` `                  ``2` `* node + ``1``); ` `  `  `    ``// Stores smallest element in ` `    ``// subarray { arr[mid + 1], arr[end] } ` `    ``int` `Y = build(tree, A, mid + ``1``, ` `                 ``end, ``2` `* node + ``2``); ` `  `  `    ``// Stores smallest element in ` `    ``// subarray { arr[start], arr[end] } ` `    ``return` `(tree[node] = Math.min(X, Y)); ` `} ` `  `  `// Function to find the smallest ` `// element present in a subarray ` `static` `int` `query(``int` `tree[], ``int` `start, ``int` `end, ` `                 ``int` `l, ``int` `r, ``int` `node) ` `{ ` `     `  `    ``// If elements of the subarray ` `    ``// are not in the range [l, r] ` `    ``if` `(start > r || end < l) ` `        ``return` `Integer.MAX_VALUE; ` `  `  `    ``// If all the elements of the ` `    ``// subarray are in the range [l, r] ` `    ``if` `(start >= l && end <= r) ` `        ``return` `tree[node]; ` `  `  `    ``// Divide tree into left ` `    ``// and right subtree ` `    ``int` `mid = (start + end) / ``2``; ` `  `  `    ``// Stores smallest element ` `    ``// in left subtree ` `    ``int` `X = query(tree, start, mid, l, ` `                  ``r, ``2` `* node + ``1``); ` `  `  `    ``// Stores smallest element in ` `    ``// right subtree ` `    ``int` `Y = query(tree, mid + ``1``, end, l, ` `                  ``r, ``2` `* node + ``2``); ` `  `  `    ``return` `Math.min(X, Y); ` `} ` `  `  `// Function that find length of longest ` `// subarray with all equal elements in ` `// atmost K decrements ` `static` `int` `longestSubArray(``int``[] A, ``int` `N, ``int` `K) ` `{ ` `     `  `    ``// Stores length of longest subarray ` `    ``// with all equal elements in atmost ` `    ``// K decrements. ` `    ``int` `res = ``1``; ` `  `  `    ``// Store the prefix sum array ` `    ``int` `preSum[] = ``new` `int``[N + ``1``]; ` `  `  `    ``// Calculate the prefix sum array ` `    ``preSum[``0``] = A[``0``]; ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `        ``preSum[i + ``1``] = preSum[i] + A[i]; ` `  `  `    ``int` `tree[] = ``new` `int``[``4` `* N + ``5``]; ` `  `  `    ``// Build the segment tree ` `    ``// for range min query ` `    ``build(tree, A, ``0``, N - ``1``, ``0``); ` `  `  `    ``// Traverse the array ` `    ``for``(``int` `i = ``0``; i < N; i++)  ` `    ``{ ` `         `  `        ``// Stores start index ` `        ``// of the subarray ` `        ``int` `start = i; ` `  `  `        ``// Stores end index ` `        ``// of the subarray ` `        ``int` `end = N - ``1``; ` `  `  `        ``int` `mid; ` `  `  `        ``// Stores end index of ` `        ``// the longest subarray ` `        ``int` `max_index = i; ` `  `  `        ``// Performing the binary search ` `        ``// to find the endpoint ` `        ``// for the selected range ` `        ``while` `(start <= end) ` `        ``{ ` `             `  `            ``// Find the mid for binary search ` `            ``mid = (start + end) / ``2``; ` `  `  `            ``// Find the smallest element in ` `            ``// range [i, mid] using Segment Tree ` `            ``int` `min_element = query(tree, ``0``, N - ``1``, ` `                                    ``i, mid, ``0``); ` `  `  `            ``// Stores total sum of subarray ` `            ``// after K decrements ` `            ``int` `expected_sum = (mid - i + ``1``) * ` `                                ``min_element; ` `  `  `            ``// Stores sum of elements of ` `            ``// subarray before K decrements ` `            ``int` `actual_sum = preSum[mid + ``1``] -  ` `                             ``preSum[i]; ` `  `  `            ``// If subarray found with ` `            ``// all equal elements ` `            ``if` `(actual_sum - expected_sum <= K)  ` `            ``{ ` `                 `  `                ``// Update start ` `                ``start = mid + ``1``; ` `  `  `                ``// Update max_index ` `                ``max_index = Math.max(max_index, mid); ` `            ``} ` `  `  `            ``// If false, it means that ` `            ``// the selected range is invalid ` `            ``else`  `            ``{ ` `                 `  `                ``// Update end ` `                ``end = mid - ``1``; ` `            ``} ` `        ``} ` `  `  `        ``// Store the length of longest subarray ` `        ``res = Math.max(res, max_index - i + ``1``); ` `    ``} ` `  `  `    ``// Return result ` `    ``return` `res; ` `} ` `  `  `// Driver Code ` `static` `public` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = { ``1``, ``7``, ``3``, ``4``, ``5``, ``6` `}; ` `    ``int` `k = ``6``; ` `    ``int` `n = ``6``; ` `     `  `    ``System.out.print(longestSubArray(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by sanjoy_62`

Python3

 `# Python3 program to implement ` `# the above approach ` `import` `sys ` ` `  `# Function to construct Segment Tree ` `# to return the minimum element in a range ` `def` `build(tree, A, start, end, node): ` `     `  `    ``# If leaf nodes of ` `    ``# the tree are found ` `    ``if` `(start ``=``=` `end): ` `         `  `        ``# Update the value in segment ` `        ``# tree from given array ` `        ``tree[node] ``=` `A[start] ` `  `  `        ``return` `tree[node] ` `     `  `    ``# Divide left and right subtree ` `    ``mid ``=` `(``int``)((start ``+` `end) ``/` `2``) ` `  `  `    ``# Stores smallest element in ` `    ``# subarray : arr[start], arr[mid]  ` `    ``X ``=` `build(tree, A, start, mid, ` `              ``2` `*` `node ``+` `1``) ` ` `  `    ``# Stores smallest element in ` `    ``# subarray : arr[mid + 1], arr[end]  ` `    ``Y ``=` `build(tree, A, mid ``+` `1``, ` `             ``end, ``2` `*` `node ``+` `2``) ` `  `  `    ``# Stores smallest element in ` `    ``# subarray : arr[start], arr[end]  ` `    ``return` `(tree[node] ``=``=` `min``(X, Y)) ` ` `  `# Function to find the smallest ` `# element present in a subarray ` `def` `query(tree, start, end, l, r, node): ` `               `  `    ``# If elements of the subarray ` `    ``# are not in the range [l, r] ` `    ``if` `(start > r ``or` `end < l) : ` `        ``return` `sys.maxsize ` `  `  `    ``# If all the elements of the ` `    ``# subarray are in the range [l, r] ` `    ``if` `(start >``=` `l ``and` `end <``=` `r): ` `        ``return` `tree[node] ` `  `  `    ``# Divide tree into left ` `    ``# and right subtree ` `    ``mid ``=` `(``int``)((start ``+` `end) ``/` `2``) ` `  `  `    ``# Stores smallest element ` `    ``# in left subtree ` `    ``X ``=` `query(tree, start, mid, l, ` `              ``r, ``2` `*` `node ``+` `1``) ` `  `  `    ``# Stores smallest element in ` `    ``# right subtree ` `    ``Y ``=` `query(tree, mid ``+` `1``, end, l, ` `            ``r, ``2` `*` `node ``+` `2``) ` `  `  `    ``return` `min``(X, Y) ` ` `  `# Function that find length of longest ` `# subarray with all equal elements in ` `# atmost K decrements ` `def` `longestSubArray(A, N, K): ` `     `  `    ``# Stores length of longest subarray ` `    ``# with all equal elements in atmost ` `    ``# K decrements. ` `    ``res ``=` `1` `  `  `    ``# Store the prefix sum array ` `    ``preSum ``=` `[``0``] ``*` `(N ``+` `1``) ` `  `  `    ``# Calculate the prefix sum array ` `    ``preSum[``0``] ``=` `A[``0``] ` `    ``for` `i ``in` `range``(N): ` `        ``preSum[i ``+` `1``] ``=` `preSum[i] ``+` `A[i] ` `  `  `    ``tree ``=` `[``0``] ``*` `(``4` `*` `N ``+` `5``) ` `  `  `    ``# Build the segment tree ` `    ``# for range min query ` `    ``build(tree, A, ``0``, N ``-` `1``, ``0``) ` `  `  `    ``# Traverse the array ` `    ``for` `i ``in` `range``(N): ` `  `  `        ``# Stores start index ` `        ``# of the subarray ` `        ``start ``=` `i ` `  `  `        ``# Stores end index ` `        ``# of the subarray ` `        ``end ``=` `N ``-` `1` `  `  `        ``# Stores end index of ` `        ``# the longest subarray ` `        ``max_index ``=` `i ` `  `  `        ``# Performing the binary search ` `        ``# to find the endpoint ` `        ``# for the selected range ` `        ``while` `(start <``=` `end): ` `  `  `            ``# Find the mid for binary search ` `            ``mid ``=` `(``int``)((start ``+` `end) ``/` `2``) ` `  `  `            ``# Find the smallest element in ` `            ``# range [i, mid] using Segment Tree ` `            ``min_element ``=` `query(tree, ``0``, N ``-` `1``, i, mid, ``0``) ` `  `  `            ``# Stores total sum of subarray ` `            ``# after K decrements ` `            ``expected_sum ``=` `(mid ``-` `i ``+` `1``) ``*` `min_element ` `  `  `            ``# Stores sum of elements of ` `            ``# subarray before K decrements ` `            ``actual_sum ``=` `preSum[mid ``+` `1``] ``-` `preSum[i] ` `  `  `            ``# If subarray found with ` `            ``# all equal elements ` `            ``if` `(actual_sum ``-` `expected_sum <``=` `K): ` `                 `  `                ``# Update start ` `                ``start ``=` `mid ``+` `1` `  `  `                ``# Update max_index ` `                ``max_index ``=` `max``(max_index, mid) ` `             `  `            ``# If false, it means that ` `            ``# the selected range is invalid ` `            ``else``: ` `  `  `                ``# Update end ` `                ``end ``=` `mid ``-` `1` `             `  `        ``# Store the length of longest subarray ` `        ``res ``=` `max``(res, max_index ``-` `i ``+` `2``) ` ` `  `    ``# Return result ` `    ``return` `res ` ` `  `# Driver Code ` `arr ``=` `[ ``1``, ``7``, ``3``, ``4``, ``5``, ``6` `] ` `k ``=` `6` `n ``=` `6` ` `  `print``(longestSubArray(arr, n, k)) ` ` `  `# This code is contributed by splevel62`

C#

 `// C# program to implement  ` `// the above approach  ` `using` `System; ` `  `  `class` `GFG{ ` `      `  `// Function to construct Segment Tree ` `// to return the minimum element in a range ` `static` `int` `build(``int``[] tree, ``int``[] A, ``int` `start, ` `                 ``int` `end, ``int` `node) ` `{ ` `     `  `    ``// If leaf nodes of ` `    ``// the tree are found ` `    ``if` `(start == end)  ` `    ``{ ` `         `  `        ``// Update the value in segment ` `        ``// tree from given array ` `        ``tree[node] = A[start]; ` `   `  `        ``return` `tree[node]; ` `    ``} ` `   `  `    ``// Divide left and right subtree ` `    ``int` `mid = (start + end) / 2; ` `   `  `    ``// Stores smallest element in ` `    ``// subarray { arr[start], arr[mid] } ` `    ``int` `X = build(tree, A, start, mid, ` `                  ``2 * node + 1); ` `   `  `    ``// Stores smallest element in ` `    ``// subarray { arr[mid + 1], arr[end] } ` `    ``int` `Y = build(tree, A, mid + 1, ` `                  ``end, 2 * node + 2); ` `   `  `    ``// Stores smallest element in ` `    ``// subarray { arr[start], arr[end] } ` `    ``return` `(tree[node] = Math.Min(X, Y)); ` `} ` `   `  `// Function to find the smallest ` `// element present in a subarray ` `static` `int` `query(``int``[] tree, ``int` `start, ``int` `end, ` `                 ``int` `l, ``int` `r, ``int` `node) ` `{ ` `      `  `    ``// If elements of the subarray ` `    ``// are not in the range [l, r] ` `    ``if` `(start > r || end < l) ` `        ``return` `Int32.MaxValue; ` `   `  `    ``// If all the elements of the ` `    ``// subarray are in the range [l, r] ` `    ``if` `(start >= l && end <= r) ` `        ``return` `tree[node]; ` `   `  `    ``// Divide tree into left ` `    ``// and right subtree ` `    ``int` `mid = (start + end) / 2; ` `   `  `    ``// Stores smallest element ` `    ``// in left subtree ` `    ``int` `X = query(tree, start, mid, l, ` `                  ``r, 2 * node + 1); ` `   `  `    ``// Stores smallest element in ` `    ``// right subtree ` `    ``int` `Y = query(tree, mid + 1, end, l, ` `                  ``r, 2 * node + 2); ` `   `  `    ``return` `Math.Min(X, Y); ` `} ` `   `  `// Function that find length of longest ` `// subarray with all equal elements in ` `// atmost K decrements ` `static` `int` `longestSubArray(``int``[] A, ``int` `N, ``int` `K) ` `{ ` `      `  `    ``// Stores length of longest subarray ` `    ``// with all equal elements in atmost ` `    ``// K decrements. ` `    ``int` `res = 1; ` `   `  `    ``// Store the prefix sum array ` `    ``int``[] preSum = ``new` `int``[N + 1]; ` `   `  `    ``// Calculate the prefix sum array ` `    ``preSum[0] = A[0]; ` `    ``for``(``int` `i = 0; i < N; i++) ` `        ``preSum[i + 1] = preSum[i] + A[i]; ` `   `  `    ``int``[] tree = ``new` `int``[4 * N + 5]; ` `   `  `    ``// Build the segment tree ` `    ``// for range min query ` `    ``build(tree, A, 0, N - 1, 0); ` `   `  `    ``// Traverse the array ` `    ``for``(``int` `i = 0; i < N; i++)  ` `    ``{ ` `          `  `        ``// Stores start index ` `        ``// of the subarray ` `        ``int` `start = i; ` `   `  `        ``// Stores end index ` `        ``// of the subarray ` `        ``int` `end = N - 1; ` `   `  `        ``int` `mid; ` `   `  `        ``// Stores end index of ` `        ``// the longest subarray ` `        ``int` `max_index = i; ` `   `  `        ``// Performing the binary search ` `        ``// to find the endpoint ` `        ``// for the selected range ` `        ``while` `(start <= end) ` `        ``{ ` `              `  `            ``// Find the mid for binary search ` `            ``mid = (start + end) / 2; ` `   `  `            ``// Find the smallest element in ` `            ``// range [i, mid] using Segment Tree ` `            ``int` `min_element = query(tree, 0, N - 1, ` `                                    ``i, mid, 0); ` `   `  `            ``// Stores total sum of subarray ` `            ``// after K decrements ` `            ``int` `expected_sum = (mid - i + 1) * ` `                                ``min_element; ` `   `  `            ``// Stores sum of elements of ` `            ``// subarray before K decrements ` `            ``int` `actual_sum = preSum[mid + 1] -  ` `                             ``preSum[i]; ` `   `  `            ``// If subarray found with ` `            ``// all equal elements ` `            ``if` `(actual_sum - expected_sum <= K)  ` `            ``{ ` `                  `  `                ``// Update start ` `                ``start = mid + 1; ` `   `  `                ``// Update max_index ` `                ``max_index = Math.Max(max_index, mid); ` `            ``} ` `   `  `            ``// If false, it means that ` `            ``// the selected range is invalid ` `            ``else` `            ``{ ` `                  `  `                ``// Update end ` `                ``end = mid - 1; ` `            ``} ` `        ``} ` `   `  `        ``// Store the length of longest subarray ` `        ``res = Math.Max(res, max_index - i + 1); ` `    ``} ` `   `  `    ``// Return result ` `    ``return` `res; ` `} ` `  `  `// Driver Code ` `static` `void` `Main() ` `{ ` `    ``int``[] arr = { 1, 7, 3, 4, 5, 6 }; ` `    ``int` `k = 6; ` `    ``int` `n = 6; ` `      `  `    ``Console.WriteLine(longestSubArray(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by susmitakundugoaldanga`

Javascript

 ``

Output:

`4`

Time Complexity: O(N * (log(N))2)
Auxiliary Space: O(N)

Related Topic: Segment Tree

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next