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# Reduce the max of Array by breaking array parts at most Okay occasions

Given an integer array arr[] of dimension N and a constructive integer Okay, the duty is to attenuate the utmost of the array by changing any component arr[i] into two constructive parts (X, Y) at most Okay occasions such that arr[i] = X + Y.

Examples:

Enter: arr = {9}, Okay = 2
Output: 3
Rationalization: Operation 1: Substitute component 9 into {6, 3} then array turns into {6, 3}.
Operation 2: Substitute component 6 into {3, 3} then array turns into {3, 3, 3}.
So, the utmost component in arr[] after acting at most Okay operations are 3.

Enter: arr = {2, 4, 8, 2}, Okay = 4
Output: 2

The issue may be solved utilizing binary search based mostly on the next thought:

Initilze begin with minimal potential reply and finish with most potential reply, then calculate the edge worth mid = (begin + finish) /2 and test whether it is potential to make each component lower than or equals to mid in at most Okay operations. Whether it is potential, replace the consequence and shift finish of vary to mid – 1. In any other case, shift begin of vary to mid + 1.

Comply with the steps under to implement the above thought:

• Initialize a variable begin = 1 and finish = most potential reply.
• Initialize a variable consequence that can retailer the reply
• Whereas begin ≤ finish
• Calculate mid = (begin + finish) / 2
• Calculate the utmost variety of operations required to make each component lower than or equal to mid.
• Iterate over the given array
• Verify if the present component is bigger than mid
• If true, then calculate the operation required to make this component lower than or equals to mid
• Verify if the full operation required to make each component lower than or equal to mid is bigger lower than equal to Okay
• If true, replace the consequence and transfer the finish to mid – 1
• In any other case, transfer the begin to mid + 1.
• Return the consequence.

Beneath is the implementation of the above strategy.

## C++

 ` `  `#embrace ` `utilizing` `namespace` `std;` ` `  `int` `minimizeMaxElement(vector<``int``>& arr, ``int` `Okay)` `{` `    ` `    ` `    ``int` `begin = 1,` `        ``finish = *max_element(arr.start(), arr.finish());` ` `  `    ` `    ` `    ``int` `consequence = -1;` ` `  `    ` `    ``whereas` `(begin <= finish) {` ` `  `        ` `        ``int` `mid = (begin + finish) >> 1;` ` `  `        ` `        ` `        ` `        ` `        ``int` `operation = 0;` ` `  `        ` `        ``for` `(``int` `i = 0; i < arr.dimension(); i++) {` ` `  `            ` `            ` `            ` `            ` `            ` `            ``if` `(arr[i] > mid) {` `                ``operation += ``ceil``((``double``)arr[i] / mid) - 1;` `            ``}` `        ``}` ` `  `        ` `        ` `        ` `        ` `        ` `        ` `        ``if` `(operation <= Okay) {` `            ``consequence = mid;` `            ``finish = mid - 1;` `        ``}` `        ``else` `{` `            ``begin = mid + 1;` `        ``}` `    ``}` ` `  `    ` `    ``return` `consequence;` `}` ` `  `int` `most important()` `{` ` `  `    ``vector<``int``> arr = { 2, 4, 8, 2 };` `    ``int` `Okay = 4;` ` `  `    ` `    ``cout << minimizeMaxElement(arr, Okay);` ` `  `    ``return` `0;` `}`

Time Complexity: O(log2(max(arr)) * N), the place max(arr) is the utmost component and N is the scale of the given array.
Auxiliary Area: O(1)

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