Minimum Operations to Form Subsequence With Target Sum

Minimum Operations to Form Subsequence With Target Sum - Problem

You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target.

In one operation, you must apply the following changes to the array:

Choose any element of the array nums[i] such that nums[i] > 1.
Remove nums[i] from the array.
Add two occurrences of nums[i] / 2 to the end of nums.

Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target. If it is impossible to obtain such a subsequence, return -1.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Input & Output

Example 1 — Basic Split Required

$ Input: nums = [1,1,4], target = 3

› Output: 1

💡 Note: We have two 1s and one 4. To form target 3, we need a 2. We can split 4 → [2,2] in 1 operation, then use 1+2=3.

Example 2 — No Operations Needed

$ Input: nums = [1,2,8], target = 3

› Output: 0

💡 Note: We already have 1 and 2 in the array. We can directly form 1+2=3 without any operations.

Example 3 — Impossible Case

$ Input: nums = [1,2], target = 4

› Output: -1

💡 Note: Total sum is 1+2=3, which is less than target 4. It's impossible to reach target 4.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 10⁶
nums[i] is a power of 2
1 ≤ target ≤ 10⁶

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to use greedy bit manipulation - process target bit by bit and split larger powers of 2 only when needed. The optimal approach counts frequencies and uses binary representation to minimize operations. Time: O(log target), Space: O(log max(nums)).

Common Approaches

✓ Brute Force Simulation

⏱️ Time: O(2^n) Space: O(2^n)

Generate all possible states by applying operations recursively, then check if any state contains a subsequence summing to target. This explores the entire state space.

Greedy Bit Manipulation

⏱️ Time: O(log target) Space: O(log max(nums))

Count frequency of each power of 2, then greedily form the target by using largest denominations first. Split larger denominations only when needed.

Brute Force Simulation — Algorithm Steps

Recursively try all possible operations on each element > 1
For each state, check if subsequence sum equals target
Return minimum operations among valid states

Visualization

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Step-by-Step Walkthrough

Initial State

Start with given array of powers of 2

Try Splits

Recursively try splitting each element > 1 into two halves

Check Target

For each state, check if subsequence can sum to target

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

#define MAX_VAL 1048576
int available[MAX_VAL + 1];

int canFormTarget(int target) {
    int remaining = target;
    for (int power = 20; power >= 0; power--) {
        int val = 1 << power;
        if (val <= MAX_VAL) {
            int use = (available[val] < remaining / val) ? available[val] : remaining / val;
            remaining -= use * val;
        }
    }
    return remaining == 0;
}

int solve(int target, int ops) {
    if (canFormTarget(target)) {
        return ops;
    }
    
    int result = INT_MAX;
    for (int val = 2; val <= MAX_VAL; val <<= 1) {
        if (available[val] > 0) {
            available[val]--;
            int halfVal = val / 2;
            available[halfVal] += 2;
            
            int subResult = solve(target, ops + 1);
            if (subResult < result) {
                result = subResult;
            }
            
            // Backtrack
            available[halfVal] -= 2;
            available[val]++;
        }
    }
    
    return result;
}

int solution(int* nums, int numsSize, int target) {
    memset(available, 0, sizeof(available));
    long long sum = 0;
    
    for (int i = 0; i < numsSize; i++) {
        available[nums[i]]++;
        sum += nums[i];
    }
    
    if (sum < target) return -1;
    
    int result = solve(target, 0);
    return result == INT_MAX ? -1 : result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    
    char* ptr = strchr(line, '[');
    ptr++;
    char* token = strtok(ptr, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int target;
    scanf("%d", &target);
    
    printf("%d\n", solution(nums, numsSize, target));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Each element can be split recursively, creating exponential states

⚠ Quadratic Growth

Space Complexity

O(2^n)

Recursive call stack and state storage grows exponentially

⚠ Quadratic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

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