**Maximize the Minimum Value After K Sublist Increments- Google Top Interview Questions**

### Problem Statement :

You are given a list of integers nums and integers size and k. Consider an operation where we take a contiguous sublist of length size and increment every element by one. Given that you can perform this operation k times, return the largest minimum value possible in nums. Constraints n ≤ 100,000 where n is the length of nums k < 2 ** 31 Example 1 Input nums = [1, 4, 1, 1, 6] size = 3 k = 2 Output 2 Explanation We can increment [1, 4, 1] to get [2, 5, 2, 1, 6] and then increment [5, 2, 1] to get [2, 6, 3, 2, 6].

### Solution :

` ````
Solution in C++ :
#define ll long long
bool check(ll mid, vector<int>& a, int size, int k) {
ll n = a.size();
ll prefix[n];
memset(prefix, 0, sizeof(prefix));
ll sum = 0;
ll moves = 0;
for (ll i = 0; i < n; i++) {
sum = sum + prefix[i];
ll change = mid - (a[i] + sum);
if (change > 0) {
moves += change;
sum = sum + change;
if (i + size < n) prefix[i + size] -= change;
}
}
return moves <= k;
}
int solve(vector<int>& a, int size, int k) {
int n = a.size();
ll mi = (*min_element(a.begin(), a.end()));
ll low = mi;
ll high = mi + k;
ll ans = mi;
while (low <= high) {
ll mid = low + (high - low) / 2;
bool x = check(mid, a, size, k);
if (x == true) {
ans = max(ans, mid);
low = mid + 1;
} else {
high = mid - 1;
}
}
return (int)ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int size, int k) {
int l = 0;
int r = Integer.MAX_VALUE;
int res = -1;
while (l <= r) {
int m = l + (r - l) / 2;
if (possible(nums, size, k, m)) {
res = Math.max(res, m);
l = m + 1;
} else {
r = m - 1;
}
}
return res;
}
public boolean possible(int[] nums, int size, int k, int min) {
int n = nums.length;
int[] dec = new int[n + 1];
int increase = 0;
for (int i = 0; i < n; i++) {
increase -= dec[i];
if (increase + nums[i] < min) {
int diff = min - (increase + nums[i]);
k -= diff;
if (k < 0)
return false;
increase += diff;
if (i + size < n)
dec[i + size] += diff;
}
}
return true;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, A, size, K):
N = len(A)
def possible(target):
events = [0] * N
moves = s = 0
for i in range(N):
s += events[i]
delta = target - (A[i] + s)
if delta > 0:
moves += delta
s += delta
if i + size < N:
events[i + size] -= delta
return moves <= K
lo, hi = 0, 10 ** 10
while lo < hi:
mi = lo + hi + 1 >> 1
if possible(mi):
lo = mi
else:
hi = mi - 1
return lo
```

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