# Minimum length subarray containing all unique elements after Q operations

Given an array of size N containing all elements as 0 initially, and a Q queries containing range in the form of [L, R]. The task is to modify the array by adding 1 to each element in the range [L, R] for Q queries and then print the size of minimum length subarray containing all unique elements.

Note: 1-based indexing is used in range [L, R].

Examples:

Input: N = 6, Q[] = {{1, 3}, {4, 6}, {3, 4}, {3, 3}}
Output: 3
Explanation:
Initial array: arr[] = { 0, 0, 0, 0, 0, 0 }
Query 1: updated arr[] = { 1, 1, 1, 0, 0, 0 }.
Query 2: updated arr[] = { 1, 1, 1, 1, 1, 1 }.
Query 3: updated arr[] = { 1, 1, 2, 2, 1, 1 }.
Query 4: updated arr[] = { 1, 1, 3, 2, 1, 1 }.
The subarray { 1, 3, 2 } is minimum subarray which contains all unique elements. Thus, the answer is 3.

Input: N = 8, Q[] = {{1, 4}, {3, 4}, {4, 5}, {5, 5}, {7, 8}, {8, 8}}
Output: 4
Explanation:
After processing all queries, the array becomes = { 1, 1, 2, 3, 2, 0, 1, 2 }.
The subarray { 3, 2, 0, 1 } is minimum subarray which contains all unique elements. Thus, the answer is 4.

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

Approach: The idea is to use the concept of Prefix sum array and Two pointers approach for this problem.

• Final Array after the queries can be computed by incrementing the value at the array by 1 at the index L and decrementing the value by 1 at the index R + 1.
```Processing of a Query:
arr[L] += 1
arr[R + 1] -= 1
```
• Finally, compute the Prefix sum of the array to compute the final array after the queries. Then use Two pointers approach to get the minimum length subarray containing all unique elements.

Below is the implementation of above approach:

## C++

 `// C++ implementation to find the ` `// minimum size subarray containing ` `// all unique elements after processing ` `// the array for K queries of ranges ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find minimum size subarray ` `// of all array elements ` `int` `subarrayLength(``int` `A[], ``int` `R[], ` `                   ``int` `N, ``int` `M) ` `{ ` ` `  `    ``// Updating the array after ` `    ``// processing each query ` `    ``for` `(``int` `i = 0; i < M; ++i) { ` ` `  `        ``int` `l = R[i], r = R[i] + 1; ` ` `  `        ``// Making it to 0-indexing ` `        ``l--; ` `        ``r--; ` ` `  `        ``// Prefix sum array concept is used ` `        ``// to obtain the count array ` `        ``A[l]++; ` ` `  `        ``if` `(r < N) ` `            ``A[r]--; ` `    ``} ` ` `  `    ``// Iterating over the array ` `    ``// to get the final array ` `    ``for` `(``int` `i = 1; i < N; ++i) { ` `        ``A[i] += A[i - 1]; ` `    ``} ` ` `  `    ``// Variable to get count ` `    ``// of all unique elements ` `    ``int` `count = 0; ` ` `  `    ``// Hash to maintain perviously ` `    ``// occured elements ` `    ``unordered_set<``int``> s; ` ` `  `    ``// Loop to find the all ` `    ``// unique elements ` `    ``for` `(``int` `i = 0; i < N; ++i) { ` `        ``if` `(s.find(A[i]) == s.end()) ` `            ``count++; ` ` `  `        ``s.insert(A[i]); ` `    ``} ` ` `  `    ``// array to maintain counter ` `    ``// of encountered elements ` `    ``vector<``int``> repeat(count + 1, 0); ` ` `  `    ``// variable to store answer ` `    ``int` `ans = N; ` ` `  `    ``// Using two pointers approach ` `    ``int` `counter = 0, left = 0, right = 0; ` ` `  `    ``while` `(right < N) { ` ` `  `        ``int` `cur_element = A[right]; ` `        ``repeat[cur_element] += 1; ` ` `  `        ``// Increment counter ` `        ``// if occured once ` `        ``if` `(repeat[cur_element] == 1) ` `            ``++counter; ` ` `  `        ``// when all unique ` `        ``// elements are found ` `        ``while` `(counter == count) { ` ` `  `            ``// update answer with ` `            ``// minimum size ` `            ``ans = min(ans, right - left + 1); ` ` `  `            ``// decrement count of ` `            ``// elements from left ` `            ``cur_element = A[left]; ` `            ``repeat[cur_element] -= 1; ` `            ``++left; ` ` `  `            ``// decrement counter ` `            ``if` `(repeat[cur_element] == 0) ` `                ``--counter; ` `        ``} ` ` `  `        ``++right; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 8, queries = 6; ` `    ``int` `Q[] ` `        ``= { ` `            ``{ 1, 4 }, { 3, 4 }, { 4, 5 }, ` `            ``{ 5, 5 }, { 7, 8 }, { 8, 8 } ` `          ``}; ` ` `  `    ``int` `A[N] = { 0 }; ` ` `  `    ``cout << subarrayLength(A, Q, N, queries); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to find the ` `# minimum size subarray containing ` `# all unique elements after processing ` `# the array for K queries of ranges ` ` `  `# Function to find minimum size subarray ` `# of all array elements ` `def` `subarrayLength(A, R, N, M): ` ` `  `    ``# Updating the array after ` `    ``# processing each query ` `    ``for` `i ``in` `range``(M): ` ` `  `        ``l ``=` `R[i][``0``] ` `        ``r ``=` `R[i][``1``] ``+` `1` ` `  `        ``# Making it to 0-indexing ` `        ``l ``-``=` `1` `        ``r ``-``=` `1` ` `  `        ``# Prefix sum array concept is used ` `        ``# to obtain the count array ` `        ``A[l] ``+``=` `1` ` `  `        ``if` `(r < N): ` `            ``A[r] ``-``=` `1` ` `  `    ``# Iterating over the array ` `    ``# to get the final array ` `    ``for` `i ``in` `range``(``1` `, N): ` `        ``A[i] ``+``=` `A[i ``-` `1``] ` ` `  `    ``# Variable to get count ` `    ``# of all unique elements ` `    ``count ``=` `0` ` `  `    ``# Hash to maintain perviously ` `    ``# occured elements ` `    ``s ``=` `[] ` ` `  `    ``# Loop to find the all ` `    ``# unique elements ` `    ``for` `i ``in` `range``(N): ` `        ``if` `(A[i] ``not` `in` `s): ` `            ``count ``+``=` `1` ` `  `        ``s.append(A[i]) ` ` `  `    ``# Array to maintain counter ` `    ``# of encountered elements ` `    ``repeat ``=` `[``0``] ``*` `(count ``+` `1``) ` ` `  `    ``# Variable to store answer ` `    ``ans ``=` `N ` ` `  `    ``# Using two pointers approach ` `    ``counter, left, right ``=` `0``, ``0``, ``0` ` `  `    ``while` `(right < N): ` ` `  `        ``cur_element ``=` `A[right] ` `        ``repeat[cur_element] ``+``=` `1` ` `  `        ``# Increment counter ` `        ``# if occured once ` `        ``if` `(repeat[cur_element] ``=``=` `1``): ` `            ``counter ``+``=` `1` ` `  `        ``# When all unique ` `        ``# elements are found ` `        ``while` `(counter ``=``=` `count): ` ` `  `            ``# update answer with ` `            ``# minimum size ` `            ``ans ``=` `min``(ans, right ``-` `left ``+` `1``) ` ` `  `            ``# Decrement count of ` `            ``# elements from left ` `            ``cur_element ``=` `A[left] ` `            ``repeat[cur_element] ``-``=` `1` `            ``left ``+``=` `1` ` `  `            ``# Decrement counter ` `            ``if` `(repeat[cur_element] ``=``=` `0``): ` `                ``counter ``-``=` `1` `                 `  `        ``right ``+``=` `1` `         `  `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``N , queries ``=` `8` `, ``6` `    ``Q ``=` `[ [ ``1``, ``4` `], [ ``3``, ``4` `], [ ``4``, ``5` `], ` `          ``[ ``5``, ``5` `], [ ``7``, ``8` `], [ ``8``, ``8` `] ] ` ` `  `    ``A ``=` `[``0``] ``*` `N ` `    ``print``(subarrayLength(A, Q, N, queries)) ` ` `  `# This code is contributed by chitranayal `

Output:

```4
```

Time Complexity: O(N)

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Improved By : chitranayal