@@ -88,32 +88,295 @@ tags:
8888
8989<!-- solution:start -->
9090
91- ### 方法一
91+ ### 方法一:贪心 + 前缀和 + 二分查找
92+ 93+ 我们考虑枚举 Alice 的站立位置 $i,ドル对于每个 $i,ドル我们按照如下的策略进行操作:
94+ 95+ - 首先,如果位置 $i$ 的数字为 1ドル,ドル我们可以直接拾取一个 1ドル,ドル不需要行动次数。
96+ - 然后,我们对 $i$ 的左右两侧位置的数字 1ドル$ 进行拾取,执行的是行动 2ドル,ドル即把位置 $i-1$ 的 1ドル$ 移到位置 $i,ドル然后拾取;把位置 $i+1$ 的 1ドル$ 移到位置 $i,ドル然后拾取。每拾取一个 1ドル,ドル需要 1ドル$ 次行动。
97+ - 接下来,我们最大限度地将 $i-1$ 或 $i+1$ 上的 0ドル,ドル利用行动 1ドル,ドル将其置为 1ドル,ドル然后利用行动 2ドル,ドル将其移动到位置 $i,ドル拾取。直到拾取的 1ドル$ 的数量达到 $k$ 或者行动 1ドル$ 的次数达到 $\text{maxChanges}$。我们假设行动 1ドル$ 的次数为 $c,ドル那么总共需要 2ドルc$ 次行动。
98+ - 利用完行动 1ドル,ドル如果拾取的 1ドル$ 的数量还没有达到 $k,ドル我们需要继续考虑在 $[ 1,..i-2] $ 和 $[ i+2,..n] $ 的区间内,进行行动 2ドル,ドル将 1ドル$ 移动到位置 $i,ドル拾取。我们可以使用二分查找来确定这个区间的大小,使得拾取的 1ドル$ 的数量达到 $k$。具体地,我们二分枚举一个区间的大小 $d,ドル然后在区间 $[ i-d,..i-2] $ 和 $[ i+2,..i+d] $ 内,进行行动 2ドル,ドル将 1ドル$ 移动到位置 $i,ドル拾取。如果拾取的 1ドル$ 的数量达到 $k,ドル我们就更新答案。
99+ 100+ 时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $\text{nums}$ 的长度。
92101
93102<!-- tabs:start -->
94103
95104#### Python3
96105
97106``` python
98- 107+ class Solution :
108+ def minimumMoves (self nums : List[int ], k : int , maxChanges : int ) -> int :
109+ n = len (nums)
110+ cnt = [0 ] * (n + 1 )
111+ s = [0 ] * (n + 1 )
112+ for i, x in enumerate (nums, 1 ):
113+ cnt[i] = cnt[i - 1 ] + x
114+ s[i] = s[i - 1 ] + i * x
115+ ans = inf
116+ max = lambda x , y : x if x > y else y
117+ min = lambda x , y : x if x < y else y
118+ for i, x in enumerate (nums, 1 ):
119+ t = 0
120+ need = k - x
121+ for j in (i - 1 , i + 1 ):
122+ if need > 0 and 1 <= j <= n and nums[j - 1 ] == 1 :
123+ need -= 1
124+ t += 1
125+ c = min (need, maxChanges)
126+ need -= c
127+ t += c * 2
128+ if need <= 0 :
129+ ans = min (ans, t)
130+ continue
131+ l, r = 2 , max (i - 1 , n - i)
132+ while l <= r:
133+ mid = (l + r) >> 1
134+ l1, r1 = max (1 , i - mid), max (0 , i - 2 )
135+ l2, r2 = min (n + 1 , i + 2 ), min (n, i + mid)
136+ c1 = cnt[r1] - cnt[l1 - 1 ]
137+ c2 = cnt[r2] - cnt[l2 - 1 ]
138+ if c1 + c2 >= need:
139+ t1 = c1 * i - (s[r1] - s[l1 - 1 ])
140+ t2 = s[r2] - s[l2 - 1 ] - c2 * i
141+ ans = min (ans, t + t1 + t2)
142+ r = mid - 1
143+ else :
144+ l = mid + 1
145+ return ans
99146```
100147
101148#### Java
102149
103150``` java
104- 151+ class Solution {
152+ public long minimumMoves (int [] nums , int k , int maxChanges ) {
153+ int n = nums. length;
154+ int [] cnt = new int [n + 1 ];
155+ long [] s = new long [n + 1 ];
156+ for (int i = 1 ; i <= n; ++ i) {
157+ cnt[i] = cnt[i - 1 ] + nums[i - 1 ];
158+ s[i] = s[i - 1 ] + i * nums[i - 1 ];
159+ }
160+ long ans = Long . MAX_VALUE
161+ for (int i = 1 ; i <= n; ++ i) {
162+ long t = 0 ;
163+ int need = k - nums[i - 1 ];
164+ for (int j = i - 1 ; j <= i + 1 ; j += 2 ) {
165+ if (need > 0 && 1 <= j && j <= n && nums[j - 1 ] == 1 ) {
166+ -- need;
167+ ++ t;
168+ }
169+ }
170+ int c = Math . min(need, maxChanges);
171+ need -= c;
172+ t += c * 2 ;
173+ if (need <= 0 ) {
174+ ans = Math . min(ans, t);
175+ continue ;
176+ }
177+ int l = 2 , r = Math . max(i - 1 , n - i);
178+ while (l <= r) {
179+ int mid = (l + r) >> 1 ;
180+ int l1 = Math . max(1 , i - mid), r1 = Math . max(0 , i - 2 );
181+ int l2 = Math . min(n + 1 , i + 2 ), r2 = Math . min(n, i + mid);
182+ int c1 = cnt[r1] - cnt[l1 - 1 ];
183+ int c2 = cnt[r2] - cnt[l2 - 1 ];
184+ if (c1 + c2 >= need) {
185+ long t1 = 1L * c1 * i - (s[r1] - s[l1 - 1 ]);
186+ long t2 = s[r2] - s[l2 - 1 ] - 1L * c2 * i;
187+ ans = Math . min(ans, t + t1 + t2);
188+ r = mid - 1 ;
189+ } else {
190+ l = mid + 1 ;
191+ }
192+ }
193+ }
194+ return ans;
195+ }
196+ }
105197```
106198
107199#### C++
108200
109201``` cpp
110- 202+ class Solution {
203+ public:
204+ long long minimumMoves(vector<int >& nums, int k, int maxChanges) {
205+ int n = nums.size();
206+ vector<int > cnt(n + 1, 0);
207+ vector<long long > s(n + 1, 0);
208+ 209+ for (int i = 1; i <= n; ++i) {
210+ cnt[i] = cnt[i - 1] + nums[i - 1];
211+ s[i] = s[i - 1] + 1LL * i * nums[i - 1];
212+ }
213+ 214+ long long ans = LLONG_MAX;
215+ 216+ for (int i = 1 ; i <= n; ++i) {
217+ long long t = 0;
218+ int need = k - nums[i - 1];
219+ 220+ for (int j = i - 1; j <= i + 1; j += 2) {
221+ if (need > 0 && 1 <= j && j <= n && nums[j - 1] == 1) {
222+ --need;
223+ ++t;
224+ }
225+ }
226+ 227+ int c = min(need, maxChanges);
228+ need -= c;
229+ t += c * 2;
230+ 231+ if (need <= 0) {
232+ ans = min(ans, t);
233+ continue;
234+ }
235+ 236+ int l = 2, r = max(i - 1, n - i);
237+ 238+ while (l <= r) {
239+ int mid = (l + r) / 2;
240+ int l1 = max(1, i - mid), r1 = max(0, i - 2);
241+ int l2 = min(n + 1, i + 2), r2 = min(n, i + mid);
242+ 243+ int c1 = cnt[r1] - cnt[l1 - 1];
244+ int c2 = cnt[r2] - cnt[l2 - 1];
245+ 246+ if (c1 + c2 >= need) {
247+ long long t1 = 1LL * c1 * i - (s[r1] - s[l1 - 1]);
248+ long long t2 = s[r2] - s[l2 - 1] - 1LL * c2 * i;
249+ ans = min(ans, t + t1 + t2);
250+ r = mid - 1;
251+ } else {
252+ l = mid + 1;
253+ }
254+ }
255+ }
256+ 257+ return ans;
258+ }
259+ };
111260```
112261
113262#### Go
114263
115264``` go
265+ func minimumMoves (nums []int , k int , maxChanges int ) int64 {
266+ n := len (nums)
267+ cnt := make ([]int , n+1 )
268+ s := make ([]int , n+1 )
269+ 270+ for i := 1 ; i <= n; i++ {
271+ cnt[i] = cnt[i-1 ] + nums[i-1 ]
272+ s[i] = s[i-1 ] + i*nums[i-1 ]
273+ }
274+ 275+ ans := math.MaxInt64
276+ 277+ for i := 1 ; i <= n; i++ {
278+ t := 0
279+ need := k - nums[i-1 ]
280+ 281+ for _ , j := range []int {i - 1 , i + 1 } {
282+ if need > 0 && 1 <= j && j <= n && nums[j-1 ] == 1 {
283+ need--
284+ t++
285+ }
286+ }
287+ 288+ c := min (need, maxChanges)
289+ need -= c
290+ t += c * 2
291+ 292+ if need <= 0 {
293+ ans = min (ans, t)
294+ continue
295+ }
296+ 297+ l , r := 2 , max (i-1 , n-i)
298+ 299+ for l <= r {
300+ mid := (l + r) >> 1
301+ l1 , r1 := max (1 , i-mid), max (0 , i-2 )
302+ l2 , r2 := min (n+1 , i+2 ), min (n, i+mid)
303+ 304+ c1 := cnt[r1] - cnt[l1-1 ]
305+ c2 := cnt[r2] - cnt[l2-1 ]
306+ 307+ if c1+c2 >= need {
308+ t1 := c1*i - (s[r1] - s[l1-1 ])
309+ t2 := s[r2] - s[l2-1 ] - c2*i
310+ ans = min (ans, t+t1+t2)
311+ r = mid - 1
312+ } else {
313+ l = mid + 1
314+ }
315+ }
316+ }
317+ 318+ return int64 (ans)
319+ }
320+ ```
116321
322+ #### TypeScript
323+ 324+ ``` ts
325+ function minimumMoves(nums : number [], k : number , maxChanges : number ): number {
326+ const = nums .length ;
327+ const = Array (n + 1 ).fill (0 );
328+ const = Array (n + 1 ).fill (0 );
329+ 330+ for (let i = 1 ; i <= n ; i ++ ) {
331+ cnt [i ] = cnt [i - 1 ] + nums [i - 1 ];
332+ s [i ] = s [i - 1 ] + i * nums [i - 1 ];
333+ }
334+ 335+ let ans = Infinity ;
336+ for (let i = 1 ; i <= n ; i ++ ) {
337+ let t = 0 ;
338+ let need = k - nums [i - 1 ];
339+ 340+ for (let j of [i - 1 , i + 1 ]) {
341+ if (need > 0 && 1 <= j && j <= n && nums [j - 1 ] === 1 ) {
342+ need -- ;
343+ t ++ ;
344+ }
345+ }
346+ 347+ const = Math .min (need , maxChanges );
348+ need -= c ;
349+ t += c * 2 ;
350+ 351+ if (need <= 0 ) {
352+ ans = Math .min (ans , t );
353+ continue ;
354+ }
355+ 356+ let l = 2 ,
357+ r = Math .max (i - 1 , n - i );
358+ 359+ while (l <= r ) {
360+ const = (l + r ) >> 1 ;
361+ const = [Math .max (1 , i - mid ), Math .max (0 , i - 2 )];
362+ const = [Math .min (n + 1 , i + 2 ), Math .min (n , i + mid )];
363+ 364+ const = cnt [r1 ] - cnt [l1 - 1 ];
365+ const = cnt [r2 ] - cnt [l2 - 1 ];
366+ 367+ if (c1 + c2 >= need ) {
368+ const = c1 * i - (s [r1 ] - s [l1 - 1 ]);
369+ const = s [r2 ] - s [l2 - 1 ] - c2 * i ;
370+ ans = Math .min (ans , t + t1 + t2 );
371+ r = mid - 1 ;
372+ } else {
373+ l = mid + 1 ;
374+ }
375+ }
376+ }
377+ 378+ return ans ;
379+ }
117380```
118381
119382<!-- tabs:end -->
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