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Most worth you possibly can get hold of at index 0 by performing given operation

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Given an array nums[] of dimension N, Discover the most worth you possibly can obtain at index 0 after performing the operation the place in every operation enhance the worth at index 0 by 1 and reduce the worth at index i by 1 such that nums[0] < nums[i] (1 ≤ i ≤ N-1). You may carry out as many strikes as you want to (probably, zero).


Enter: nums[] = {1, 2, 3}
Output: 3
Rationalization: nums[0] < nums[1], Due to this fact nums[0] + 1 and nums[1] – 1. Now,  nums[0] = 2, nums[0] < nums[2], once more repeat the step. Now nums[0] turns into 3 which is the utmost potential worth we are able to get at index 0.

Enter: nums[] = {1, 2, 2}
Output: 2

Strategy: The issue will be solved based mostly on the next thought:

With a view to obtain most worth at index 0, kind the array from index 1 to the final index and also can begin iterating it from index 1 and if at any level nums[i] is discovered to be better than the ingredient current at index 0, in line with the given operation nums[0] get elevated and nums[i] get decreased. So, nums[0] get elevated by the quantity (nums[i] – nums[0] + 1) / 2 when encountered with a better worth.

Comply with the steps talked about beneath to implement the concept:

  • Retailer the preliminary worth current at index 0 in a variable.
  • Type the array from index 1 to the final index.
  • Iterate from index 1 and if discovered nums[i] > nums[0], do the given operation
  • Worth at index 0 will get elevated by the (nums[i] – nums[0] +1) / 2
  • Return the worth at index 0 as the required reply.

Beneath is the implementation of the above method:



#embrace <bits/stdc++.h>

utilizing namespace std;


int maximumValue(vector<int>& nums, int n)



    int value_at_index_0 = nums[0];




    kind(nums.start() + 1, nums.finish());



    for (int i = 1; i < n; i++) {





        if (nums[i] > value_at_index_0) {




                += (nums[i] - value_at_index_0 + 1) / 2;





    return value_at_index_0;



int principal()


    vector<int> nums = { 1, 2, 3 };

    int N = nums.dimension();



    cout << maximumValue(nums, N);


    return 0;


Time Complexity: O(N * log N)
Auxiliary Area: O(1)

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