LeetCode //C - 964. Least Operators to Express Number
964. Least Operators to Express Number
Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x … where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, *, or /). For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.
When writing such an expression, we adhere to the following conventions:
- The division operator (/) returns rational numbers.
- There are no parentheses placed anywhere.
- We use the usual order of operations: multiplication and division happen before addition and subtraction.
- It is not allowed to use the unary negation operator (-). For example, “x - x” is a valid expression as it only uses subtraction, but “-x + x” is not because it uses negation.
We would like to write an expression with the least number of operators such that the expression equals the given target. Return the least number of operators used.
Example 1:
Input: x = 3, target = 19
Output: 5
Explanation: 3 * 3 + 3 * 3 + 3 / 3.
The expression contains 5 operations.
Example 2:
Input: x = 5, target = 501
Output: 8
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.
The expression contains 8 operations.
Example 3:
Input: x = 100, target = 100000000
Output: 3
Explanation: 100 * 100 * 100 * 100.
The expression contains 3 operations.
Constraints:
- 2 <= x <= 100
- 1 < = t a r g e t < = 2 ∗ 10 8 1 <= target <= 2 * 10^8 1<=target<=2∗108
From: LeetCode
Link: 964. Least Operators to Express Number
Solution:
Ideas:
- For small values v <= x, directly compute the best using only 1s (x/x) or as x minus some 1s.
- For larger values, find the closest power x^k and try:
- Undershoot: use x^(k-1) and build the remainder.
- Overshoot: use x^k and subtract the difference (only if it helps).
- Use memoization to avoid recomputing subproblems.
Code:
typedefstruct{int key;int val;} Pair;static Pair memo[10000];staticint memoSize;staticintdfs(int x,int v){// Base case: v <= xif(v <= x){// Option 1: v = 1 + 1 + ... + 1 (v times)// 1 is x / x -> 1 operator// plus (v - 1) additions// total = v (division) + (v - 1) (+) = 2*v - 1int op_add =2* v -1;// Option 2: v = x - ( (x - v) ones )// (x - v) ones: 2*(x - v) - 1 operators// one more '-' to subtract from x// total = 2*(x - v)int op_sub =2*(x - v);return op_add < op_sub ? op_add : op_sub;}// Check memofor(int i =0; i < memoSize;++i){if(memo[i].key == v)return memo[i].val;}// Find smallest k such that x^k >= vint k =2;long y =(long)x * x;// y = x^2 initiallywhile(y < v){ y *= x;++k;// now y = x^k}// Now y = x^k >= v, and y/x = x^(k-1)// Option 1 (undershoot):// Use x^(k-1) once (cost k-1 multiplications),// then express the remaining (v - x^(k-1))int op1 =(k -1)+dfs(x, v -(int)(y / x));int ans = op1;// Option 2 (overshoot), only if the "over" part is smaller than v:// Use x^k once (cost k multiplications),// then express (x^k - v), and subtract it.if(y - v < v){int op2 = k +dfs(x,(int)(y - v));if(op2 < ans) ans = op2;}// Save to memo memo[memoSize].key = v; memo[memoSize].val = ans;++memoSize;return ans;}intleastOpsExpressTarget(int x,int target){ memoSize =0;// reset memo for each test casereturndfs(x, target);}
