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1 | 1 | package g3301_3400.s3395_subsequences_with_a_unique_middle_mode_i;
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2 | 2 |
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3 |
| -// #Hard #Dynamic_Programming #Sliding_Window #Combinatorics #Subsequence |
4 |
| -// #2024_12_22_Time_1115_ms_(100.00%)_Space_45.2_MB_(100.00%) |
| 3 | +// #Hard #Array #Hash_Table #Math #Combinatorics #2025_01_06_Time_27_(99.29%)_Space_45.15_(97.87%) |
5 | 4 |
|
6 |
| -import java.util.ArrayList; |
7 | 5 | import java.util.HashMap;
|
8 |
| -import java.util.List; |
9 | 6 | import java.util.Map;
|
10 | 7 |
|
11 | 8 | public class Solution {
|
12 |
| - private static final int MOD = 1000000007; |
| 9 | + private static final int MOD = (int) 1e9 + 7; |
| 10 | + private long[] c2 = new long[1001]; |
13 | 11 |
|
14 |
| - public int subsequencesWithMiddleMode(int[] a) { |
15 |
| - int n = a.length; |
16 |
| - // Create a dictionary to store indices of each number |
17 |
| - Map<Integer, List<Integer>> dict = new HashMap<>(); |
18 |
| - for (int i = 0; i < n; i++) { |
19 |
| - dict.computeIfAbsent(a[i], k -> new ArrayList<>()).add(i); |
20 |
| - } |
21 |
| - long ans = 0L; |
22 |
| - // Iterate over each unique number and its indices |
23 |
| - for (Map.Entry<Integer, List<Integer>> entry : dict.entrySet()) { |
24 |
| - List<Integer> b = entry.getValue(); |
25 |
| - int m = b.size(); |
26 |
| - for (int k = 0; k < m; k++) { |
27 |
| - int i = b.get(k); |
28 |
| - int r = m - 1 - k; |
29 |
| - int u = i - k; |
30 |
| - int v = (n - 1 - i) - r; |
31 |
| - // Case 2: Frequency of occurrence is 2 times |
32 |
| - ans = (ans + convert(k, 1) * convert(u, 1) % MOD * convert(v, 2) % MOD) % MOD; |
33 |
| - ans = (ans + convert(r, 1) * convert(u, 2) % MOD * convert(v, 1) % MOD) % MOD; |
34 |
| - // Case 3: Frequency of occurrence is 3 times |
35 |
| - ans = (ans + convert(k, 2) * convert(v, 2) % MOD) % MOD; |
36 |
| - ans = (ans + convert(r, 2) * convert(u, 2) % MOD) % MOD; |
37 |
| - ans = |
38 |
| - (ans |
39 |
| - + convert(k, 1) |
40 |
| - * convert(r, 1) |
41 |
| - % MOD |
42 |
| - * convert(u, 1) |
43 |
| - % MOD |
44 |
| - * convert(v, 1) |
45 |
| - % MOD) |
46 |
| - % MOD; |
47 |
| - |
48 |
| - // Case 4: Frequency of occurrence is 4 times |
49 |
| - ans = (ans + convert(k, 2) * convert(r, 1) % MOD * convert(v, 1) % MOD) % MOD; |
50 |
| - ans = (ans + convert(k, 1) * convert(r, 2) % MOD * convert(u, 1) % MOD) % MOD; |
51 |
| - |
52 |
| - // Case 5: Frequency of occurrence is 5 times |
53 |
| - ans = (ans + convert(k, 2) * convert(r, 2) % MOD) % MOD; |
| 12 | + public int subsequencesWithMiddleMode(int[] nums) { |
| 13 | + if (c2[2] == 0) { |
| 14 | + c2[0] = c2[1] = 0; |
| 15 | + c2[2] = 1; |
| 16 | + for (int i = 3; i < c2.length; ++i) { |
| 17 | + c2[i] = i * (i - 1) / 2; |
54 | 18 | }
|
55 | 19 | }
|
56 |
| - long dif = 0; |
57 |
| - // Principle of inclusion-exclusion |
58 |
| - for (Map.Entry<Integer, List<Integer>> midEntry : dict.entrySet()) { |
59 |
| - List<Integer> b = midEntry.getValue(); |
60 |
| - int m = b.size(); |
61 |
| - for (Map.Entry<Integer, List<Integer>> tmpEntry : dict.entrySet()) { |
62 |
| - if (!midEntry.getKey().equals(tmpEntry.getKey())) { |
63 |
| - List<Integer> c = tmpEntry.getValue(); |
64 |
| - int size = c.size(); |
65 |
| - int k = 0; |
66 |
| - int j = 0; |
67 |
| - while (k < m) { |
68 |
| - int i = b.get(k); |
69 |
| - int r = m - 1 - k; |
70 |
| - int u = i - k; |
71 |
| - int v = (n - 1 - i) - r; |
72 |
| - while (j < size && c.get(j) < i) { |
73 |
| - j++; |
74 |
| - } |
75 |
| - int x = j; |
76 |
| - int y = size - x; |
77 |
| - dif = |
78 |
| - (dif |
79 |
| - + convert(k, 1) |
80 |
| - * convert(x, 1) |
81 |
| - % MOD |
82 |
| - * convert(y, 1) |
83 |
| - % MOD |
84 |
| - * convert(v - y, 1) |
85 |
| - % MOD) |
86 |
| - % MOD; |
87 |
| - dif = |
88 |
| - (dif |
89 |
| - + convert(k, 1) |
90 |
| - * convert(y, 2) |
91 |
| - % MOD |
92 |
| - * convert(u - x, 1) |
93 |
| - % MOD) |
94 |
| - % MOD; |
95 |
| - dif = |
96 |
| - (dif + convert(k, 1) * convert(x, 1) % MOD * convert(y, 2) % MOD) |
97 |
| - % MOD; |
98 |
| - |
99 |
| - dif = |
100 |
| - (dif |
101 |
| - + convert(r, 1) |
102 |
| - * convert(x, 1) |
103 |
| - % MOD |
104 |
| - * convert(y, 1) |
105 |
| - % MOD |
106 |
| - * convert(u - x, 1) |
107 |
| - % MOD) |
108 |
| - % MOD; |
109 |
| - dif = |
110 |
| - (dif |
111 |
| - + convert(r, 1) |
112 |
| - * convert(x, 2) |
113 |
| - % MOD |
114 |
| - * convert(v - y, 1) |
115 |
| - % MOD) |
116 |
| - % MOD; |
117 |
| - dif = |
118 |
| - (dif + convert(r, 1) * convert(x, 2) % MOD * convert(y, 1) % MOD) |
119 |
| - % MOD; |
120 |
| - k++; |
121 |
| - } |
122 |
| - } |
| 20 | + int n = nums.length; |
| 21 | + int[] newNums = new int[n]; |
| 22 | + Map<Integer, Integer> map = new HashMap<>(n); |
| 23 | + int m = 0; |
| 24 | + int index = 0; |
| 25 | + for (int x : nums) { |
| 26 | + Integer id = map.get(x); |
| 27 | + if (id == null) { |
| 28 | + id = m++; |
| 29 | + map.put(x, id); |
123 | 30 | }
|
| 31 | + newNums[index++] = id; |
124 | 32 | }
|
125 |
| - return (int) ((ans - dif + MOD) % MOD); |
126 |
| - } |
127 |
| - |
128 |
| - private long convert(int n, int k) { |
129 |
| - if (k > n) { |
| 33 | + if (m == n) { |
130 | 34 | return 0;
|
131 | 35 | }
|
132 |
| - if (k == 0 || k == n) { |
133 |
| - return 1; |
| 36 | + int[] rightCount = new int[m]; |
| 37 | + for (int x : newNums) { |
| 38 | + rightCount[x]++; |
134 | 39 | }
|
135 |
| - long res = 1; |
136 |
| - for (int i = 0; i < k; i++) { |
137 |
| - res = res * (n - i) / (i + 1); |
| 40 | + int[] leftCount = new int[m]; |
| 41 | + long ans = (long) n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120; |
| 42 | + for (int left = 0; left < n - 2; left++) { |
| 43 | + int x = newNums[left]; |
| 44 | + rightCount[x]--; |
| 45 | + if (left >= 2) { |
| 46 | + int right = n - (left + 1); |
| 47 | + int leftX = leftCount[x]; |
| 48 | + int rightX = rightCount[x]; |
| 49 | + ans -= c2[left - leftX] * c2[right - rightX]; |
| 50 | + for (int y = 0; y < m; ++y) { |
| 51 | + if (y == x) { |
| 52 | + continue; |
| 53 | + } |
| 54 | + int rightY = rightCount[y]; |
| 55 | + int leftY = leftCount[y]; |
| 56 | + ans -= c2[leftY] * rightX * (right - rightX); |
| 57 | + ans -= c2[rightY] * leftX * (left - leftX); |
| 58 | + ans -= |
| 59 | + leftY |
| 60 | + * rightY |
| 61 | + * (leftX * (right - rightX - rightY) |
| 62 | + + rightX * (left - leftX - leftY)); |
| 63 | + } |
| 64 | + } |
| 65 | + leftCount[x]++; |
138 | 66 | }
|
139 |
| - return res % MOD; |
| 67 | + return (int) (ans % MOD); |
140 | 68 | }
|
141 | 69 | }
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