# Target Sum (Leetcode 494) - Coding Interview Question Mani
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You are given an integer array `nums` and an integer `target`.

You want to build an expression out of nums by adding one of the symbols `'+'` and `'-'` before each integer in nums and then concatenate all the integers.

• For example, if `nums = [2, 1]`, you can add a `'+'` before `2` and a `'-'` before `1` and concatenate them to build the expression `"+2-1"`.

Return the number of different expressions that you can build, which evaluates to `target`.

Example 1:

```Input: nums = [1,1,1,1,1], target = 3
Output: 5
Explanation: There are 5 ways to assign symbols to make the sum of nums be target 3.
-1 + 1 + 1 + 1 + 1 = 3
+1 - 1 + 1 + 1 + 1 = 3
+1 + 1 - 1 + 1 + 1 = 3
+1 + 1 + 1 - 1 + 1 = 3
+1 + 1 + 1 + 1 - 1 = 3
```

Example 2:

```Input: nums = , target = 1
Output: 1
```

Constraints:

• `1 <= nums.length <= 20`
• `0 <= nums[i] <= 1000`
• `0 <= sum(nums[i]) <= 1000`
• `-1000 <= target <= 1000`

Solution:

``````1
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/**
* Intuition: This is a 0/1 knapsack dp. This is equivalent to given an array of numbers, count the
* number of subsets whose sum of elements is = (arraySum + target) / 2. Here all the elemenets
* with + sign can be considered as grouped into subset1 and all elements of - sign are grouped into subset2.
* One thing to note is that if the arraySum + target is odd it means we cannot divide the array into two halfs
* with that sum.
* <p>
* Time Complexity: O(N ^ 2)
* Space Complexity: O(N ^ 2)
**/
class Solution {
public int findTargetSumWays(int[] a, int target) {
int sum = 0;
int zeros = 0;
for (int i = 0; i < a.length; i++) {
if (a[i] == 0) {
zeros++;
} else {
sum += a[i];
}
}

int[] w = new int[a.length - zeros];
int k = 0;
for (int i : a) {
if (i != 0) {
w[k++] = i;
}
}

int c = sum + target;
if (c % 2 == 1) {
return 0;
}

c = c / 2;

int[][] dp = new int[w.length + 1][c + 1];
for (int i = 0; i < dp.length; i++) {
dp[i] = 1;
}

for (int i = 1; i < dp.length; i++) {
for (int j = 1; j < dp.length; j++) {
if (w[i - 1] <= j) {
dp[i][j] = dp[i - 1][j - w[i - 1]] + dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j];
}
}
}

//         for(int d[] : dp) {
//             System.out.println(Arrays.toString(d));
//         }

return dp[w.length][c] * (int) Math.pow(2, zeros);

}
}
``````
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