Given an infinite stream of integers, find the k’th largest element at any point of time. It may be assumed that 1 <= k <= n.

Input: stream[] = {10, 20, 11, 70, 50, 40, 100, 5, ...} k = 3 Output: {_, _, 10, 11, 20, 40, 50, 50, ...}

Extra space allowed is O(k).

The idea is to use min heap.

1) Store first k elements in min heap.

2) For every element from (k+1)-th to n-th, do following.

……a) Print root of heap.

……b) If current element is more than root of heap, pop root and insert

`// CPP program to find k-th largest element in a ` `// stream after every insertion. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; `
` ` `int` `kthLargest(` `int` `stream[], ` `int` `n, ` `int` `k) `
`{ ` ` ` `// Create a min heap and store first k-1 elements `
` ` `// of stream into `
` ` `priority_queue<` `int` `, vector<` `int` `>, greater<` `int` `> > pq; `
` ` ` ` `// Push first k elements and print "_" (k-1) times `
` ` `for` `(` `int` `i=0; i<k-1; i++) `
` ` `{ `
` ` `pq.push(stream[i]); `
` ` `cout << ` `"_ "` `; `
` ` `} `
` ` `pq.push(stream[k-1]); `
` ` ` ` `for` `(` `int` `i=k; i<n; i++) `
` ` `{ `
` ` `// We must insert last element before we `
` ` `// decide last k-th largest output. `
` ` `cout << pq.top() << ` `" "` `; `
` ` ` ` `if` `(stream[i] > pq.top()) `
` ` `{ `
` ` `pq.pop(); `
` ` `pq.push(stream[i]); `
` ` `} `
` ` `} `
` ` ` ` `// Print last k-th largest element (after `
` ` `// (inserting last element) `
` ` `cout << pq.top(); `
`} ` ` ` `// Driver code ` `int` `main() `
`{ ` ` ` `int` `arr[] = {10, 20, 11, 70, 50, 40, 100, 55}; `
` ` `int` `k = 3; `
` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]); `
` ` `kthLargest(arr, n, k); `
` ` `return` `0; `
`} ` |

*chevron_right*

*filter_none*

**Output:**

_ _ 10 11 20 40 50 55

If stream contains elements of non-primitive types, we may define our own compactor function and create a priority_queue accordingly.

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