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leetcode/algorithms/cpp/patchingArray/PatchingArray.cpp at master · fallleaves/leetcode · GitHub
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PatchingArray.cpp
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//
Source : https://leetcode.com/problems/patching-array/
//
Author : Hao Chen
//
Date : 2016-03-01
/*
**************************************************************************************
*
* Given a sorted positive integer array nums and an integer n, add/patch elements to
* the array such that any number in range [1, n] inclusive can be formed by the sum of
* some elements in the array. Return the minimum number of patches required.
*
* Example 1:
* nums = [1, 3], n = 6
* Return 1.
*
* Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4.
* Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3],
* [1,2,3].
* Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6].
* So we only need 1 patch.
*
* Example 2:
* nums = [1, 5, 10], n = 20
* Return 2.
* The two patches can be [2, 4].
*
* Example 3:
* nums = [1, 2, 2], n = 5
* Return 0.
*
* Credits:Special thanks to @dietpepsi for adding this problem and creating all test
* cases.
**************************************************************************************
*/
class
Solution
{
public:
int
minPatches
(vector<
int
>& nums,
int
n) {
return
minPatches_02
(nums, n);
return
minPatches_01
(nums, n);
}
//
Greedy Algorithm
//
(Assume the array is sorted already)
//
//
Let do some observation at first,
//
//
1) if we have [1,2] then we can cover 1, 2, 3
//
2) if we have [1,2,3] then we can cover 1,2,3,4,5,6
//
So, it looks we can simply add all of nums together, then we can find out max number we can reach.
//
//
3) if we have [1,2,5], we can see
//
3.1) [1,2] can cover 1,2,3, but we cannot reach 4,
//
3.2) then we patch 4, then we have [1,2,4] which can cover 1,2,3(1+2),4,5(1+4),6(2+4), 7(1+2+4)
//
3.3) we can see [1,2,4] can reach to 7 - sum all of them
//
3.4) then [1,2,4,5], we can reach to 12 - 1,2,3,4,5,6,7,8(1+2+5),9(4+5),10(1+4+5), 11(2+4+5), 12(1+2+4+5)
//
//
So, we can have figure out our solution
//
//
0) considering the `n` we need to cover.
//
1) maintain a variable we are trying to patch, suppose named `try_patch`
//
2) if `try_patch` >= nums[i] then, we just keep add the current array item,
//
and set the `try_patch` to the next patch candidate number - `sum+1`
//
3) if `try_patch` < nums[i], which means we need to patch.
//
int
minPatches_01
(vector<
int
>& nums,
int
n) {
long
covered =
0
;
//
avoid integer overflow
int
patch_cnt =
0
;
int
i =
0
;
while
(i<nums.
size
() ){
//
set the `try_patch` is the next number which we cannot cover
int
try_patch = covered +
1
;
//
if the `try_patch` can cover the current item, then just sum it,
//
then we can have the max number we can cover so far
if
( try_patch >= nums[i]) {
covered += nums[i];
i++;
}
else
{
//
if the `try_patch` cannot cover the current item, then we find the number we need to patch
patch_cnt++;
//
cout << "patch " << try_patch << endl;
covered = covered + try_patch;
}
if
(covered >=n)
break
;
}
//
for the case - [1], 7
//
the above while-loop just process all of the numbers in the array,
//
but we might not reach the goal, so, we need keep patching.
while
(covered < n) {
int
try_patch = covered +
1
;
patch_cnt++;
//
cout << "patch " << try_patch << endl;
covered = covered + try_patch;
}
return
patch_cnt;
}
//
The following solution just re-organizes the solution above, and make it shorter
int
minPatches_02
(vector<
int
>& nums,
int
n) {
long
covered =
0
;
int
patch_cnt =
0
;
int
i =
0
;
while
( covered < n){
if
(i<nums.
size
() && nums[i] <= covered +
1
) {
covered += nums[i++];
}
else
{
//
cout << "patch " << covered + 1 << endl;
covered =
2
* covered +
1
;
patch_cnt++;
}
}
return
patch_cnt;
}
};
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