Minimum Toggle to Partition

Time: O(n)

Space: O(1)

Intuition

The implementation evaluates toggle cost by using global counts and running prefix counts while scanning possible partition points. For a split, it computes cost for the pattern left=0 and right=1: flip all 1s seen on the left plus all 0s remaining on the right. The minimum across splits (and baseline uniform cases considered in code) is returned.

Algorithm

1Count total zeros (cnt_total_0) and ones (cnt_total_1) in the whole array.
2Initialise prefix counts till_cnt_0 and till_cnt_1 as 0.
3Initialise minToggles with baseline values used in implementation.
4Scan array from left to right, updating prefix counts at each element.
5For each prefix boundary, compute toggles = till_cnt_1 + (cnt_total_0 - till_cnt_0).
6Update minToggles with the minimum observed value.
7Return minToggles.

Example Walkthrough

Input: arr = [1, 0, 1, 0]

1.Totals: zeros=2, ones=2.
2.After processing prefix [1], till_cnt_1=1, till_cnt_0=0 -> cost = 1 + (2-0) = 3.
3.After prefix [1,0], till_cnt_1=1, till_cnt_0=1 -> cost = 1 + (2-1) = 2.
4.Continue all boundaries and keep the minimum cost.

Output: 2

Common Pitfalls

•The formula shown is specifically for left=0, right=1 partition cost; changing target pattern needs a different expression.
•Be consistent with whether split is before or after current index when updating prefix counts.
•Duplicates and already-uniform arrays should still be handled by baseline minimum initialization.

Minimum Toggle to Partition.java

Java


// Approach: Count total zeros/ones, then scan once while treating each position as a partition boundary.
// Compute toggles for making left all 0s and right all 1s using prefix counts and take the minimum.
// Also compare with uniform-array conversions considered in the implementation.
// Time: O(n) Space: O(1)

class Solution {

    int minToggle(int[] arr) {
        int cnt_total_0 = 0;
        int cnt_total_1 = 0;

        for (int bit : arr) {
            if (bit == 0) {
                cnt_total_0++; 
            }else {
                cnt_total_1++;
            }
        }

        int till_cnt_0 = 0;
        int till_cnt_1 = 0;
        int minToggles = Integer.MAX_VALUE;
        //Case 1 ---> left side all 1s and right side all 1s
        minToggles = Math.min(minToggles, cnt_total_0);

        //Case 2 ---> left side all 0s and right side all 0s
        minToggles = Math.min(minToggles, cnt_total_0);

        for (int bit : arr) {
            if (bit == 0) {
                till_cnt_0++; 
            }else {
                till_cnt_1++;
            }

            //Case 3 ---> left side all 0s and right side all 1s
            minToggles = Math.min(minToggles, (till_cnt_1 + (cnt_total_0 - till_cnt_0)));
        }
        return minToggles;
    }
}

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