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1 | 1 | package g3301_3400.s3395_subsequences_with_a_unique_middle_mode_i; |
2 | 2 |
|
3 | | -// #Hard #Array #Hash_Table #Math #Combinatorics |
4 | | -// #2025_01_06_Time_1141_(39.01%)_Space_44.78_(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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