# Maximal Expression - Google Top Interview Questions

### Problem Statement :

Given a list of integers nums, return the maximal value that can be made by adding any binary operators +, -, and * between the numbers in nums as well as insert any valid brackets.

Constraints

1 ≤ n ≤ 200 where n is the length of nums

Example 1

Input

nums = [-5, -3, -8]

Output

64

Explanation

We can make the following expression: (-5 + -3) * -8

### Solution :

Solution in C++ :

int solve(vector<int>& A) {
int N = A.size();

int t[N][N][2];

for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++) t[i][j][0] = INT_MAX, t[i][j][1] = INT_MIN;

for (int i = 0; i < N; i++) t[i][i][0] = t[i][i][1] = A[i];

for (int L = 2; L <= N; L++)
for (int i = 0; i + L - 1 < N; i++) {
int j = i + L - 1;

for (int k = i; k < j; k++) {
int a = t[i][k][0];
int b = t[i][k][1];

int c = t[k + 1][j][0];
int d = t[k + 1][j][1];

vector<int> P = {a, b};
vector<int> Q = {c, d};

for (int m = 0; m < 2; m++)
for (int n = 0; n < 2; n++) {
t[i][j][0] = min(P[m] * Q[n], t[i][j][0]);
t[i][j][1] = max(P[m] * Q[n], t[i][j][1]);

t[i][j][0] = min(P[m] + Q[n], t[i][j][0]);
t[i][j][1] = max(P[m] + Q[n], t[i][j][1]);

t[i][j][0] = min(P[m] - Q[n], t[i][j][0]);
t[i][j][1] = max(P[m] - Q[n], t[i][j][1]);
}
}
}

return t[0][N - 1][1];
}

Solution in Java :

import java.util.*;

class Solution {
class Node {
int max;
int min;
public Node(int max, int min) {
this.max = max;
this.min = min;
}
}
public int solve(int[] nums) {
int len = nums.length;
Node[][] dp = new Node[len][len];

for (int i = len - 1; i >= 0; i--) {
for (int j = i; j < len; j++) {
dp[i][j] = new Node(Integer.MIN_VALUE, Integer.MAX_VALUE);
if (i == j) {
dp[i][j] = new Node(nums[i], nums[i]);
} else {
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;

for (int index = i; index < j; index++) {
max = Math.max(max, dp[i][index].max + dp[index + 1][j].max);
max = Math.max(max, dp[i][index].max - dp[index + 1][j].min);
max = Math.max(max,
Math.max(dp[i][index].max * dp[index + 1][j].max,
dp[i][index].min * dp[index + 1][j].min));
min = Math.min(min, dp[i][index].min + dp[index + 1][j].min);
min = Math.min(min, dp[i][index].min - dp[index + 1][j].max);
min = Math.min(min,
Math.min(dp[i][index].max * dp[index + 1][j].min,
dp[i][index].min * dp[index + 1][j].max));
}
dp[i][j] = new Node(max, min);
}
}
}

return dp[0][len - 1].max;
}
/**
HashMap<String, int[]> cache;
public int solve(int[] nums) {
int len = nums.length;
cache = new HashMap<>();
return process(nums, 0, len - 1)[0];
}

//[MAX, MIN]
public int[] process(int[] nums, int i, int j) {
//System.out.println("I" + i + "j" + j);
// if (i > j) {
//     return new int[]{Integer.MIN_VALUE, Integer.MAX_VALUE};
// }
// if (i >= nums.length || j >= nums.length) {
//     return new int[]{Integer.MIN_VALUE, Integer.MAX_VALUE};
// }
if (i == j) {
return new int[]{nums[i], nums[i]};
}

String key = i + "#" + j;
if (cache.containsKey(key)) {
return cache.get(key);
}

int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
for (int index = i; index < j; index++) {
int[] res1 = process(nums, i, index);
int[] res2 = process(nums, index + 1, j);
max = Math.max(max, res1[0] + res2[0]);
max = Math.max(max, res1[0] - res2[1]);
max = Math.max(max, Math.max(res1[0] * res2[0], res1[1] * res2[1]));
min = Math.min(min, res1[1] + res2[1]);
min = Math.min(min, res1[1] - res2[0]);
min = Math.min(min, Math.min(res1[0] * res2[1], res1[1] * res2[0]));
}
int[] res = new int[]{max, min};
cache.put(key, res);
return res;
}
*/
}

Solution in Python :

class Solution:
def solve(self, nums):
@cache
def dp(i, j):
if i == j:
return nums[i], nums[i]
elif i + 1 == j:
return max(nums[i] * nums[j], nums[i] + nums[j], nums[i] - nums[j]), min(
nums[i] * nums[j], nums[i] + nums[j], nums[i] - nums[j]
)
else:
maxx = -1e9
minn = -maxx
for k in range(i, j):
a, b = dp(i, k)
c, d = dp(k + 1, j)
maxx = max(maxx, a * c, b * d, a + c, a - d)
minn = min(minn, a * d, b * c, b + d, b - c)
return maxx, minn

a, b = dp(0, len(nums) - 1)
return a

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