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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. #define int              long long int
  4. #define double           long double
  5. #define print(a)         for(auto x : a) cout << x << " "; cout << endl
  6.  
  7.  
  8. const int M = 1000000007;
  9. const int N = 3e5+9;
  10. const int INF = 2e9+1;
  11. const int LINF = 2000000000000000001;
  12.  
  13. inline int power(int a, int b, int mod=M) {
  14.     int x = 1;
  15.     a %= mod;
  16.     while (b) {
  17.         if (b & 1) x = (x * a) % mod; 
  18.         a = (a * a) % mod;
  19.         b >>= 1;
  20.     }
  21.     return x;
  22. }
  23.  
  24.  
  25. //_ ***************************** START Below *******************************
  26.  
  27.  
  28.  
  29.  
  30. string a;
  31.  
  32. int consistency1(int n){
  33. 	vector<char> order = {'a','e','i','o','u'};
  34.  
  35.  
  36. 	vector<vector<int>> dp(n+1, vector<int>(5, 0)); // to maximize , set to null == 0 (in case of length)
  37. 	for(int i=0; i<5; i++) dp[0][i] = 0;
  38.  
  39. 	for(int i=1; i<=n; i++){
  40.  
  41. 		//* for 'a' case 
  42. 		if(a[i-1] == 'a'){
  43. 			dp[i][0] = dp[i-1][0]+1;
  44. 		}
  45. 		else{
  46. 			dp[i][0] = dp[i-1][0];
  47. 		}
  48.  
  49. 		//* for 'e' , 'i' , 'o', 'u' cases 
  50. 		for(int k=1; k<5; k++){
  51.  
  52. 			int ct = 0;
  53. 			if(a[i-1] == order[k]) ct = 1;
  54.  
  55. 			int j = i-1;
  56. 			while(j >= 0){
  57. 			    if (dp[j][k-1] > 0) dp[i][k] = max(dp[i][k], dp[j][k-1] + ct);
  58. 			    if (dp[j][k] > 0)   dp[i][k] = max(dp[i][k], dp[j][k] + ct);
  59.  
  60. 			    if (ct == 0) {
  61. 			        dp[i][k] = max(dp[i][k], dp[j][k-1]);
  62. 			        dp[i][k] = max(dp[i][k], dp[j][k]);
  63. 			    }
  64. 			    j--;
  65. 			}
  66. 		}
  67. 	}
  68.  
  69. 	return dp[n][4];
  70.  
  71. }
  72.  
  73.  
  74.  
  75.  
  76.  
  77.  
  78. int consistency2(int n){
  79.  
  80. 	vector<vector<int>> dp(n+1, vector<int>(5, 0)); // to maximize , set to null == 0 (in case of length)
  81. 	for(int i=0; i<5; i++) dp[0][i] = 0;
  82.  
  83. 	vector<char> order = {'a','e','i','o','u'};
  84.  
  85. 	for(int i=1; i<=n; i++){
  86.  
  87. 		//* for 'a' case 
  88. 		if(a[i-1] == 'a')	dp[i][0] = dp[i-1][0]+1;
  89. 		else	dp[i][0] = dp[i-1][0];
  90.  
  91. 		//* for 'e', 'i', 'o', 'u' case 
  92. 		for(int k=1; k<5; k++){
  93. 			if (a[i-1] == order[k]) {
  94. 			    if (dp[i-1][k-1] > 0 || dp[i-1][k] > 0) {
  95. 			        dp[i][k] = max(dp[i-1][k-1], dp[i-1][k]) + 1;
  96. 			    }
  97. 			} else {
  98. 			    dp[i][k] = dp[i-1][k];
  99. 			}
  100. 		}
  101. 	}
  102.  
  103. 	return dp[n][4];
  104.  
  105. }
  106.  
  107.  
  108.  
  109.  
  110.  
  111.  
  112.  
  113.  
  114.  
  115.  
  116.  
  117.  
  118.  
  119.  
  120.  
  121.  
  122.  
  123.  
  124. int practice(int n){
  125.  
  126.  
  127.     return 0;
  128. }
  129.  
  130.  
  131.  
  132.  
  133.  
  134. void solve() {
  135.  
  136.     int n;
  137.  
  138. 	cin >> a;
  139. 	n = a.size();
  140.  
  141.     cout << consistency1(n) << " " << consistency2(n) << endl;
  142.  
  143.  
  144. }
  145.  
  146.  
  147.  
  148.  
  149.  
  150. int32_t main() {
  151.     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  152.  
  153.     int t = 1;
  154.     // cin >> t;
  155.     while (t--) {
  156.         solve();
  157.     }
  158.  
  159.     return 0;
  160. }
Success #stdin #stdout 0s 5320KB
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https://ideone.com/O2coPT
language:
C++14 (gcc 8.3)
created:
1 day ago
visibility:
public

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