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describe:
There is a continuous row of houses along the street. There is a certain amount of cash hidden in each house. Now a thief plans to steal cash from these homes.
Thieves will not be able to steal from neighboring houses because they have interconnected anti-theft systems .
A thief's stealing power is defined as the maximum amount of money he can steal from a single house during the theft process .
You are given an array of integers nums representing the amount of cash stored in each house. Formally, the first i house from the left contains nums[i] U.S. dollars.
You are also given an integer that represents the minimumk number of houses that a thief will steal . A thief can always steal at least one house.k
Returns the thief's minimum stealing ability.
Example 1:
Input: nums = [2,3,5,9], k = 2 Output: 5 Explanation: The thief steals at least 2 houses, there are 3 ways: - Steal the houses at index 0 and 2, the stealing ability is max( nums[0], nums[2]) = 5. - Steal the houses at index 0 and 3, the stealing ability is max(nums[0], nums[3]) = 9. - Steal the houses at subscripts 1 and 3, the stealing ability is max(nums[1], nums[3]) = 9. Therefore, min(5, 9, 9) = 5 is returned.
Example 2:
Input: nums = [2,7,9,3,1], k = 2 Output: 2 Explanation: There are 7 ways to steal. The minimum stealing capability corresponds to stealing houses at subscripts 0 and 4. Return max(nums[0], nums[4]) = 2.
hint:
1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= (nums.length + 1)/2
Problem-solving ideas:
* Problem-solving ideas:
* Binary search method, first find the middle and see if the middle supports k requirements.
* To determine whether K is supported, we can try to find out the maximum number of ones smaller than middle within the range of nums, such as count.
* If count>=k, it means that middle is supported, and upper is updated; otherwise, lower is updated.
Code:
class Solution2560
{
public:
int minCapability(vector<int> &nums, int k)
{
int maxValue = *max_element(nums.begin(), nums.end());
int minValue = *min_element(nums.begin(), nums.end());
int abs = maxValue;
while (minValue <= maxValue)
{
int middle = (minValue + maxValue) / 2;
bool visted = false;
int count = 0;
for (int num : nums)
{
if (num <= middle && !visted)
{
count++;
visted = true;
}
else
{
visted = false;
}
}
if (count >= k)
{
abs = middle;
maxValue = middle - 1;
}
else
{
minValue = middle + 1;
}
}
return abs;
}
};