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| 1 | +//*************************** Approach 1 ***************** |
| 2 | +class Solution { |
| 3 | +public: |
| 4 | + int solve(vector<int>& nums, int &target, int i, int sum, unordered_map<string, int>& memo) { |
| 5 | + if (i == nums.size()) { |
| 6 | + return sum == target ? 1 : 0; |
| 7 | + } |
| 8 | + |
| 9 | + // Create a unique key for the current state |
| 10 | + string key = to_string(i) + "," + to_string(sum); |
| 11 | + |
| 12 | + // Check if the result is already computed |
| 13 | + if (memo.find(key) != memo.end()) { |
| 14 | + return memo[key]; |
| 15 | + } |
| 16 | + |
| 17 | + // Compute the result recursively |
| 18 | + int plus = solve(nums, target, i + 1, sum + nums[i], memo); |
| 19 | + int minus = solve(nums, target, i + 1, sum - nums[i], memo); |
| 20 | + |
| 21 | + // Store the result in the memo |
| 22 | + memo[key] = plus + minus; |
| 23 | + |
| 24 | + return memo[key]; |
| 25 | + } |
| 26 | + |
| 27 | + int findTargetSumWays(vector<int>& nums, int target) { |
| 28 | + unordered_map<string, int> memo; |
| 29 | + return solve(nums, target, 0, 0, memo); |
| 30 | + } |
| 31 | +}; |
| 32 | + |
| 33 | + |
| 34 | +// ************************** Approach 2 *************************************** |
| 35 | + |
| 36 | +class Solution { |
| 37 | +public: |
| 38 | + int S; |
| 39 | + int solve(vector<int>& nums, int &target, int i, int sum, vector<vector<int>>& t) { |
| 40 | + if(i == nums.size()) { |
| 41 | + return sum == target ? 1 : 0; |
| 42 | + } |
| 43 | + |
| 44 | + if(t[i][sum+S] != INT_MIN) { |
| 45 | + return t[i][sum+S]; |
| 46 | + } |
| 47 | + int plus = solve(nums, target, i+1, sum+nums[i], t); |
| 48 | + int minus = solve(nums, target, i+1, sum-nums[i], t); |
| 49 | + |
| 50 | + return t[i][sum+S] = plus+minus; |
| 51 | + } |
| 52 | + |
| 53 | + int findTargetSumWays(vector<int>& nums, int target) { |
| 54 | + int n = nums.size(); |
| 55 | + S = accumulate(begin(nums), end(nums), 0); |
| 56 | + vector<vector<int>> t(n, vector<int>(2*S+1, INT_MIN)); |
| 57 | + return solve(nums, target, 0, 0, t); |
| 58 | + } |
| 59 | +}; |
| 60 | + |
| 61 | +// *************************** Approach 3 ************************* |
| 62 | + |
| 63 | +class Solution { |
| 64 | +public: |
| 65 | + int S; |
| 66 | + int solve(vector<int>& nums, int &target, int i, int sum, vector<vector<int>>& t) { |
| 67 | + if(i == nums.size()) { |
| 68 | + return sum == target ? 1 : 0; |
| 69 | + } |
| 70 | + |
| 71 | + if(t[i][sum+S] != INT_MIN) { |
| 72 | + return t[i][sum+S]; |
| 73 | + } |
| 74 | + int plus = solve(nums, target, i+1, sum+nums[i], t); |
| 75 | + int minus = solve(nums, target, i+1, sum-nums[i], t); |
| 76 | + |
| 77 | + return t[i][sum+S] = plus+minus; |
| 78 | + } |
| 79 | + |
| 80 | + int findTargetSumWays(vector<int>& nums, int target) { |
| 81 | + int n = nums.size(); |
| 82 | + S = accumulate(begin(nums), end(nums), 0); |
| 83 | + vector<vector<int>> t(n, vector<int>(2*S+1, INT_MIN)); |
| 84 | + return solve(nums, target, 0, 0, t); |
| 85 | + } |
| 86 | +}; |
| 87 | + |
| 88 | +// ************************** Approach 3 *************************** |
| 89 | + |
| 90 | +class Solution { |
| 91 | +public: |
| 92 | + int t[21][1001]; |
| 93 | + int subsetSum(vector<int>& nums, int n, int s) { |
| 94 | + if(t[n][s] != -1) |
| 95 | + return t[n][s]; |
| 96 | + if(s == 0) |
| 97 | + return 1; |
| 98 | + if(n == 0) |
| 99 | + return 0; |
| 100 | + if(nums[n-1] == 0) |
| 101 | + return t[n][s] = subsetSum(nums, n-1, s); |
| 102 | + |
| 103 | + if(nums[n-1] <= s) |
| 104 | + return t[n][s] = subsetSum(nums, n-1, s-nums[n-1]) + subsetSum(nums, n-1, s); |
| 105 | + else |
| 106 | + return t[n][s] = subsetSum(nums, n-1, s); |
| 107 | + } |
| 108 | + |
| 109 | + int findTargetSumWays(vector<int>& nums, int target) { |
| 110 | + memset(t, -1, sizeof(t)); |
| 111 | + int sum = accumulate(begin(nums), end(nums), 0); |
| 112 | + auto lambda = [&](const int& x) { |
| 113 | + return x == 0; |
| 114 | + }; |
| 115 | + int zeros = count_if(begin(nums), end(nums), lambda); |
| 116 | + if(target > sum) |
| 117 | + return 0; |
| 118 | + |
| 119 | + if((sum-target) %2 != 0) |
| 120 | + return 0; |
| 121 | + |
| 122 | + int s1 = (sum-target)/2; |
| 123 | + return pow(2, zeros)*subsetSum(nums, nums.size(), s1); |
| 124 | + |
| 125 | + } |
| 126 | +}; |
| 127 | + |
| 128 | +//**************************** Approach 4 ****************************** |
| 129 | + |
| 130 | +class Solution { |
| 131 | +public: |
| 132 | + int subsetSum(vector<int>& nums, int s) { |
| 133 | + int n = nums.size(); |
| 134 | + vector<vector<int>> t(n+1, vector<int>(s+1)); |
| 135 | + |
| 136 | + for(int col = 0; col < s+1; col++) t[0][col] = 0; |
| 137 | + for(int row = 0; row < n+1; row++) t[row][0] = 1; |
| 138 | + |
| 139 | + for(int i = 1; i<n+1; i++) { |
| 140 | + for(int j = 1; j<s+1; j++) { |
| 141 | + if(nums[i-1] == 0) |
| 142 | + t[i][j] = t[i-1][j]; |
| 143 | + else if(nums[i-1] <= j) |
| 144 | + t[i][j] = t[i-1][j-nums[i-1]] + t[i-1][j]; |
| 145 | + else |
| 146 | + t[i][j] = t[i-1][j]; |
| 147 | + } |
| 148 | + } |
| 149 | + |
| 150 | + return t[n][s]; |
| 151 | + } |
| 152 | + |
| 153 | + int findTargetSumWays(vector<int>& nums, int target) { |
| 154 | + int sum = accumulate(begin(nums), end(nums), 0); |
| 155 | + auto lambda = [&](const int& x) { |
| 156 | + return x == 0; |
| 157 | + }; |
| 158 | + int zeros = count_if(begin(nums), end(nums), lambda); |
| 159 | + if(target > sum) |
| 160 | + return 0; |
| 161 | + |
| 162 | + if((sum-target) %2 != 0) |
| 163 | + return 0; |
| 164 | + |
| 165 | + int s1 = (sum-target)/2; |
| 166 | + /* |
| 167 | + You could also do like this : |
| 168 | + if((sum + target)%2 != 0) |
| 169 | + return 0; |
| 170 | + |
| 171 | + int s1 = (sum + target)/2; |
| 172 | + But since, target can be negative also as per Leetcode (they have recently changed the constraints), |
| 173 | + you need to do : |
| 174 | + target = abs(target); before finding s1 and the if condition above |
| 175 | + */ |
| 176 | + return pow(2, zeros)*subsetSum(nums, s1); |
| 177 | + } |
| 178 | +}; |
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