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# Rely of lacking components from 1 to most in index vary [L, R]

Given an array arr[] of size N, and an array queries[] of dimension Q, the duty is to search out the variety of lacking components from 1 to most within the vary of indices [L, R] the place L is queries[i][0] and R is queries[i][1].

Examples:

Enter: arr[] = {8, 6, 7, 7, 7}, queries = { {0, 3}, {1, 2}, {0, 2}, {1, 3}, {2, 3} }
Output: 5 5 5 5 6
Rationalization:
Most aspect for vary [0, 3] is 8 so variety of lacking components are 1, 2, 3, 4, 5.
Most aspect for vary [1, 2] is 7 so variety of lacking components are 1, 2, 3, 4, 5.
Most aspect for vary [0, 2] is 8 so variety of lacking components are 1, 2, 3, 4, 5.
Most aspect for vary [1, 3] is 7 so variety of lacking components are 1, 2, 3, 4, 5.
Most aspect for vary [2, 3] is 7 so variety of lacking components are 1, 2, 3, 4, 5, 6.

Enter: arr[] = {2, 6, 4, 9}, queries = { {0, 3}, {1, 2}, {0, 2}, {1, 3}, {2, 3} }
Output: 5 4 3 6 7
Rationalization:
Most aspect for vary [0, 3] is 9 so variety of lacking components are 1, 3, 5, 7, 8.
Most aspect for vary [1, 2] is 6 so variety of lacking components are 1, 2, 3, 5.
Most aspect for vary [0, 2] is 6 so variety of lacking components are 1, 3, 5.
Most aspect for vary [1, 3] is 9 so variety of lacking components are 1, 2, 3, 5, 7, 8.
Most aspect for vary [2, 3] is 9 so variety of lacking components are 1, 2, 3, 5, 6, 7, 8.

Strategy: One easy method is to course of each question linearly, retailer all components in a visited map and discover the utmost aspect. Then return the variety of lacking components from 1 to the utmost on that vary.

Observe the beneath steps to resolve this downside:

• Take one unordered map visited.
• Traverse queries from i = 0 to Q-1.
• Initialize one variable to retailer the utmost aspect of the given vary.
• Run once more one for loop from queries[i][0] to queries[i][1].
• Retailer the utmost aspect on this vary.
• Make the visited array 1 for every aspect visited in that vary.
• The variety of lacking components is identical as the worth of (most aspect – variety of distinctive components visited).

Under is the implementation of the above strategy.

## C++14

 ` `  `#embody ` `utilizing` `namespace` `std;` ` `  `vector<``int``> SolveQMax(``int``* arr, ``int``& n, ` `                      ``int` `queries[][2], ``int``& q)` `{` `    ` `    ``unordered_map<``int``, ``bool``> visited;` `    ``int` `maxElement;` `    ``vector<``int``> res;` ` `  `    ` `    ``for` `(``int` `i = 0; i < q; i++) {` `        ``maxElement = INT_MIN;` `        ``visited.clear();` `        ``for` `(``int` `j = queries[i][0]; ` `             ``j <= queries[i][1]; j++) {` `           `  `            ` `            ``maxElement = max(maxElement, ` `                             ``arr[j]);` `            ``visited[arr[j]] = 1;` `        ``}` ` `  `        ``res.push_back(maxElement ` `                      ``- visited.dimension());` `    ``}` `   `  `    ` `    ``return` `res` `}` ` `  `int` `most important()` `{` `    ``int` `N = 5;` `    ``int` `arr[] = { 8, 6, 7, 7, 7 };` `    ``int` `Q = 5;` `    ``int` `queries[][2] = { { 0, 3 }, { 1, 2 }, ` `                         ``{ 0, 2 }, { 1, 3 }, ` `                         ``{ 2, 3 } };` ` `  `    ` `    ``SolveQMax(arr, N, queries, Q);` `    ``return` `0;` `}`

Time Complexity: O(Q * N)
Auxiliary Area: O(N)

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